#include <bits/stdc++.h>
using namespace std;
const long long INF(0x3f3f3f3f3f3f3f3fll);
const long long inf(0x3f3f3f3f);
template <typename T>
void read(T &res) {
  bool flag = false;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
  for (res = ch - 48; isdigit(ch = getchar());
       res = (res << 1) + (res << 3) + ch - 48)
    ;
  flag && (res = -res);
}
template <typename T>
void Out(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) Out(x / 10);
  putchar(x % 10 + '0');
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long pow_mod(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long fact_pow(long long n, long long p) {
  long long res = 0;
  while (n) {
    n /= p;
    res += n;
  }
  return res;
}
long long mult(long long a, long long b, long long p) {
  a %= p;
  b %= p;
  long long r = 0, v = a;
  while (b) {
    if (b & 1) {
      r += v;
      if (r > p) r -= p;
    }
    v <<= 1;
    if (v > p) v -= p;
    b >>= 1;
  }
  return r;
}
long long quick_pow(long long a, long long b, long long p) {
  long long r = 1, v = a % p;
  while (b) {
    if (b & 1) r = mult(r, v, p);
    v = mult(v, v, p);
    b >>= 1;
  }
  return r;
}
bool CH(long long a, long long n, long long x, long long t) {
  long long r = quick_pow(a, x, n);
  long long z = r;
  for (long long i = 1; i <= t; i++) {
    r = mult(r, r, n);
    if (r == 1 && z != 1 && z != n - 1) return true;
    z = r;
  }
  return r != 1;
}
bool Miller_Rabin(long long n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (!(n & 1)) return false;
  long long x = n - 1, t = 0;
  while (!(x & 1)) {
    x >>= 1;
    t++;
  }
  srand(time(NULL));
  long long o = 8;
  for (long long i = 0; i < o; i++) {
    long long a = rand() % (n - 1) + 1;
    if (CH(a, n, x, t)) return false;
  }
  return true;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long t = x;
  x = y, y = t - (a / b) * y;
}
long long inv(long long a, long long b) {
  long long x, y;
  return exgcd(a, b, x, y), (x % b + b) % b;
}
long long crt(long long x, long long p, long long mod) {
  return inv(p / mod, mod) * (p / mod) * x;
}
long long fac(long long x, long long a, long long b) {
  if (!x) return 1;
  long long ans = 1;
  for (long long i = 1; i <= b; i++)
    if (i % a) ans *= i, ans %= b;
  ans = pow_mod(ans, x / b, b);
  for (long long i = 1; i <= x % b; i++)
    if (i % a) ans *= i, ans %= b;
  return ans * fac(x / a, a, b) % b;
}
long long C(long long n, long long m, long long a, long long b) {
  long long N = fac(n, a, b), M = fac(m, a, b), Z = fac(n - m, a, b), sum = 0,
            i;
  for (i = n; i; i = i / a) sum += i / a;
  for (i = m; i; i = i / a) sum -= i / a;
  for (i = n - m; i; i = i / a) sum -= i / a;
  return N * pow_mod(a, sum, b) % b * inv(M, b) % b * inv(Z, b) % b;
}
long long exlucas(long long n, long long m, long long p) {
  long long t = p, ans = 0, i;
  for (i = 2; i * i <= p; i++) {
    long long k = 1;
    while (t % i == 0) {
      k *= i, t /= i;
    }
    ans += crt(C(n, m, i, k), p, k), ans %= p;
  }
  if (t > 1) ans += crt(C(n, m, t, t), p, t), ans %= p;
  return ans % p;
}
long long prime[100000], minprime[100000];
void euler(long long n) {
  long long c = 0, i, j;
  for (i = 2; i <= n; i++) {
    if (!minprime[i]) prime[++c] = i, minprime[i] = i;
    for (j = 1; j <= c && i * prime[j] <= n; j++) {
      minprime[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
struct Tree {
  long long l, r, sum, lazy, maxn, minn;
} tree[2000000];
inline void push_up(long long rt) {
  tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
  tree[rt].maxn = max(tree[rt << 1].maxn, tree[rt << 1 | 1].maxn);
  tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}
inline void push_down(long long rt, long long length) {
  if (tree[rt].lazy) {
    tree[rt << 1].lazy += tree[rt].lazy;
    tree[rt << 1 | 1].lazy += tree[rt].lazy;
    tree[rt << 1].sum += (length - (length >> 1)) * tree[rt].lazy;
    tree[rt << 1 | 1].sum += (length >> 1) * tree[rt].lazy;
    tree[rt << 1].minn += tree[rt].lazy;
    tree[rt << 1 | 1].minn += tree[rt].lazy;
    tree[rt << 1].maxn += tree[rt].lazy;
    tree[rt << 1 | 1].maxn += tree[rt].lazy;
    tree[rt].lazy = 0;
  }
}
inline void build(long long l, long long r, long long rt, long long *aa) {
  tree[rt].lazy = 0;
  tree[rt].l = l;
  tree[rt].r = r;
  if (l == r) {
    tree[rt].sum = aa[l];
    tree[rt].minn = tree[rt].sum;
    tree[rt].maxn = tree[rt].sum;
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, rt << 1, aa);
  build(mid + 1, r, rt << 1 | 1, aa);
  push_up(rt);
}
inline void update_range(long long L, long long R, long long key,
                         long long rt) {
  if (tree[rt].r < L || tree[rt].l > R) return;
  if (L <= tree[rt].l && R >= tree[rt].r) {
    tree[rt].sum += (tree[rt].r - tree[rt].l + 1) * key;
    tree[rt].minn += key;
    tree[rt].maxn += key;
    tree[rt].lazy += key;
    return;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  if (L <= mid) update_range(L, R, key, rt << 1);
  if (R > mid) update_range(L, R, key, rt << 1 | 1);
  push_up(rt);
}
inline long long query_range(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].sum;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = 0;
  if (L <= mid) ans += query_range(L, R, rt << 1);
  if (R > mid) ans += query_range(L, R, rt << 1 | 1);
  return ans;
}
inline long long query_min(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].minn;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = (0x3f3f3f3f3f3f3f3fll);
  if (L <= mid) ans = min(ans, query_min(L, R, rt << 1));
  if (R > mid) ans = min(ans, query_min(L, R, rt << 1 | 1));
  return ans;
}
inline long long query_max(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].maxn;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = -(0x3f3f3f3f3f3f3f3fll);
  if (L <= mid) ans = max(ans, query_max(L, R, rt << 1));
  if (R > mid) ans = max(ans, query_max(L, R, rt << 1 | 1));
  return ans;
}
const long long N = 3e5 + 10;
long long aa[N];
struct node {
  long long val, id;
  bool operator<(node const &a) const { return val > a.val; }
} a[N], q[N];
signed main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  build(1, n, 1, aa);
  for (long long i = 1; i <= n; i++) cin >> a[i].val, a[i].id = i, q[i] = a[i];
  sort(a + 1, a + 1 + n);
  long long ans = 0, now = 1;
  for (long long i = n; i >= 1; i--) {
    while (now <= n && a[now].val >= i)
      update_range(a[now].id, a[now].id, 1, 1), now++;
    ans += query_range(1, min(q[i].val, i - 1), 1);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

One day Polycarp decided to rewatch his absolute favourite episode of well-known TV series "Tufurama". He was pretty surprised when he got results only for season 7 episode 3 with his search query of "Watch Tufurama season 3 episode 7 online full hd free". This got Polycarp confused — what if he decides to rewatch the entire series someday and won't be able to find the right episodes to watch? Polycarp now wants to count the number of times he will be forced to search for an episode using some different method.

TV series have n seasons (numbered 1 through n), the i-th season has ai episodes (numbered 1 through ai). Polycarp thinks that if for some pair of integers x and y (x < y) exist both season x episode y and season y episode x then one of these search queries will include the wrong results. Help Polycarp to calculate the number of such pairs!

Input

The first line contains one integer n (1 ≤ n ≤ 2·105) — the number of seasons.

The second line contains n integers separated by space a1, a2, ..., an (1 ≤ ai ≤ 109) — number of episodes in each season.

Output

Print one integer — the number of pairs x and y (x < y) such that there exist both season x episode y and season y episode x.

Examples

Input

5
1 2 3 4 5


Output

0


Input

3
8 12 7


Output

3


Input

3
3 2 1


Output

2

Note

Possible pairs in the second example:

  1. x = 1, y = 2 (season 1 episode 2 <image> season 2 episode 1); 
  2. x = 2, y = 3 (season 2 episode 3 <image> season 3 episode 2); 
  3. x = 1, y = 3 (season 1 episode 3 <image> season 3 episode 1). 



In the third example:

  1. x = 1, y = 2 (season 1 episode 2 <image> season 2 episode 1); 
  2. x = 1, y = 3 (season 1 episode 3 <image> season 3 episode 1). 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF(0x3f3f3f3f3f3f3f3fll);
const long long inf(0x3f3f3f3f);
template <typename T>
void read(T &res) {
  bool flag = false;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
  for (res = ch - 48; isdigit(ch = getchar());
       res = (res << 1) + (res << 3) + ch - 48)
    ;
  flag && (res = -res);
}
template <typename T>
void Out(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) Out(x / 10);
  putchar(x % 10 + '0');
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long pow_mod(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long fact_pow(long long n, long long p) {
  long long res = 0;
  while (n) {
    n /= p;
    res += n;
  }
  return res;
}
long long mult(long long a, long long b, long long p) {
  a %= p;
  b %= p;
  long long r = 0, v = a;
  while (b) {
    if (b & 1) {
      r += v;
      if (r > p) r -= p;
    }
    v <<= 1;
    if (v > p) v -= p;
    b >>= 1;
  }
  return r;
}
long long quick_pow(long long a, long long b, long long p) {
  long long r = 1, v = a % p;
  while (b) {
    if (b & 1) r = mult(r, v, p);
    v = mult(v, v, p);
    b >>= 1;
  }
  return r;
}
bool CH(long long a, long long n, long long x, long long t) {
  long long r = quick_pow(a, x, n);
  long long z = r;
  for (long long i = 1; i <= t; i++) {
    r = mult(r, r, n);
    if (r == 1 && z != 1 && z != n - 1) return true;
    z = r;
  }
  return r != 1;
}
bool Miller_Rabin(long long n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (!(n & 1)) return false;
  long long x = n - 1, t = 0;
  while (!(x & 1)) {
    x >>= 1;
    t++;
  }
  srand(time(NULL));
  long long o = 8;
  for (long long i = 0; i < o; i++) {
    long long a = rand() % (n - 1) + 1;
    if (CH(a, n, x, t)) return false;
  }
  return true;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long t = x;
  x = y, y = t - (a / b) * y;
}
long long inv(long long a, long long b) {
  long long x, y;
  return exgcd(a, b, x, y), (x % b + b) % b;
}
long long crt(long long x, long long p, long long mod) {
  return inv(p / mod, mod) * (p / mod) * x;
}
long long fac(long long x, long long a, long long b) {
  if (!x) return 1;
  long long ans = 1;
  for (long long i = 1; i <= b; i++)
    if (i % a) ans *= i, ans %= b;
  ans = pow_mod(ans, x / b, b);
  for (long long i = 1; i <= x % b; i++)
    if (i % a) ans *= i, ans %= b;
  return ans * fac(x / a, a, b) % b;
}
long long C(long long n, long long m, long long a, long long b) {
  long long N = fac(n, a, b), M = fac(m, a, b), Z = fac(n - m, a, b), sum = 0,
            i;
  for (i = n; i; i = i / a) sum += i / a;
  for (i = m; i; i = i / a) sum -= i / a;
  for (i = n - m; i; i = i / a) sum -= i / a;
  return N * pow_mod(a, sum, b) % b * inv(M, b) % b * inv(Z, b) % b;
}
long long exlucas(long long n, long long m, long long p) {
  long long t = p, ans = 0, i;
  for (i = 2; i * i <= p; i++) {
    long long k = 1;
    while (t % i == 0) {
      k *= i, t /= i;
    }
    ans += crt(C(n, m, i, k), p, k), ans %= p;
  }
  if (t > 1) ans += crt(C(n, m, t, t), p, t), ans %= p;
  return ans % p;
}
long long prime[100000], minprime[100000];
void euler(long long n) {
  long long c = 0, i, j;
  for (i = 2; i <= n; i++) {
    if (!minprime[i]) prime[++c] = i, minprime[i] = i;
    for (j = 1; j <= c && i * prime[j] <= n; j++) {
      minprime[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
struct Tree {
  long long l, r, sum, lazy, maxn, minn;
} tree[2000000];
inline void push_up(long long rt) {
  tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
  tree[rt].maxn = max(tree[rt << 1].maxn, tree[rt << 1 | 1].maxn);
  tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}
inline void push_down(long long rt, long long length) {
  if (tree[rt].lazy) {
    tree[rt << 1].lazy += tree[rt].lazy;
    tree[rt << 1 | 1].lazy += tree[rt].lazy;
    tree[rt << 1].sum += (length - (length >> 1)) * tree[rt].lazy;
    tree[rt << 1 | 1].sum += (length >> 1) * tree[rt].lazy;
    tree[rt << 1].minn += tree[rt].lazy;
    tree[rt << 1 | 1].minn += tree[rt].lazy;
    tree[rt << 1].maxn += tree[rt].lazy;
    tree[rt << 1 | 1].maxn += tree[rt].lazy;
    tree[rt].lazy = 0;
  }
}
inline void build(long long l, long long r, long long rt, long long *aa) {
  tree[rt].lazy = 0;
  tree[rt].l = l;
  tree[rt].r = r;
  if (l == r) {
    tree[rt].sum = aa[l];
    tree[rt].minn = tree[rt].sum;
    tree[rt].maxn = tree[rt].sum;
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, rt << 1, aa);
  build(mid + 1, r, rt << 1 | 1, aa);
  push_up(rt);
}
inline void update_range(long long L, long long R, long long key,
                         long long rt) {
  if (tree[rt].r < L || tree[rt].l > R) return;
  if (L <= tree[rt].l && R >= tree[rt].r) {
    tree[rt].sum += (tree[rt].r - tree[rt].l + 1) * key;
    tree[rt].minn += key;
    tree[rt].maxn += key;
    tree[rt].lazy += key;
    return;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  if (L <= mid) update_range(L, R, key, rt << 1);
  if (R > mid) update_range(L, R, key, rt << 1 | 1);
  push_up(rt);
}
inline long long query_range(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].sum;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = 0;
  if (L <= mid) ans += query_range(L, R, rt << 1);
  if (R > mid) ans += query_range(L, R, rt << 1 | 1);
  return ans;
}
inline long long query_min(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].minn;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = (0x3f3f3f3f3f3f3f3fll);
  if (L <= mid) ans = min(ans, query_min(L, R, rt << 1));
  if (R > mid) ans = min(ans, query_min(L, R, rt << 1 | 1));
  return ans;
}
inline long long query_max(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].maxn;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = -(0x3f3f3f3f3f3f3f3fll);
  if (L <= mid) ans = max(ans, query_max(L, R, rt << 1));
  if (R > mid) ans = max(ans, query_max(L, R, rt << 1 | 1));
  return ans;
}
const long long N = 3e5 + 10;
long long aa[N];
struct node {
  long long val, id;
  bool operator<(node const &a) const { return val > a.val; }
} a[N], q[N];
signed main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  build(1, n, 1, aa);
  for (long long i = 1; i <= n; i++) cin >> a[i].val, a[i].id = i, q[i] = a[i];
  sort(a + 1, a + 1 + n);
  long long ans = 0, now = 1;
  for (long long i = n; i >= 1; i--) {
    while (now <= n && a[now].val >= i)
      update_range(a[now].id, a[now].id, 1, 1), now++;
    ans += query_range(1, min(q[i].val, i - 1), 1);
  }
  cout << ans << '\n';
  return 0;
}

```