#include <bits/stdc++.h>
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define per(a, b, c) for(int a = (b); a >= (c); -- a)
using namespace std;
template <class T> void rd(T &x){
  x = 0; int f = 1; char ch = getchar();
  while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); }
  while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
}

const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
int n, m;
int s[2][maxn];
int mi[maxn << 2][2];

struct node{ int l, r;
  bool operator < (const node &b) const {
	return l == b.l ? r < b.r : l < b.l;
  }
}q[maxn];

#define lson u << 1, l, mid
#define rson u << 1 | 1, mid + 1, r

void update(int u, int l, int r, int x, int v, int op){
  mi[u][op] = min(mi[u][op], v);
  if(l == r) return ;
  int mid = l + r >> 1;
  if(x <= mid) update(lson, x, v, op);
  else update(rson, x, v, op);
}

int query(int u, int l, int r, int ql, int qr, int v, int op){
  if(ql <= l && r <= qr) return mi[u][op] + v;
  int mid = l + r >> 1, res = inf;
  if(ql <= mid) res = min(res, query(lson, ql, qr, v, op));
  if(qr > mid) res = min(res, query(rson, ql, qr, v, op));
  return res;
}

void sol(){
  int ans = s[0][1];
  update(1, 0, n, 0, s[1][1], 1);
  update(1, 0, n, 0, s[0][1], 0);
  rep(i, 1, m){
	int l = q[i].l, r = q[i].r;
	int t = query(1, 0, n, l, r, -s[1][r + 1], 1);
	t = min(t, query(1, 0, n, 0, l - 1, -s[0][l] + s[1][l] - s[1][r + 1], 0));
	update(1, 0, n, r, t + s[1][r + 1], 1);
	update(1, 0, n, r, t + s[0][r + 1], 0);
	ans = min(ans, t + s[0][r + 1]);
  }
  printf("%d\n", ans);
}

int main(){
  memset(mi, 0x3f, sizeof(mi));
  rd(n);
  rep(i, 1, n){
	int x; rd(x);
	s[1 - x][i] = 1;
  }
  per(i, n, 1) rep(j, 0, 1) s[j][i] += s[j][i + 1];
  rd(m);
  rep(i, 1, m) rd(q[i].l), rd(q[i].r);
  sort(q + 1, q + m + 1);
  sol();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include <bits/stdc++.h>
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define per(a, b, c) for(int a = (b); a >= (c); -- a)
using namespace std;
template <class T> void rd(T &x){
  x = 0; int f = 1; char ch = getchar();
  while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); }
  while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
}

const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
int n, m;
int s[2][maxn];
int mi[maxn << 2][2];

struct node{ int l, r;
  bool operator < (const node &b) const {
	return l == b.l ? r < b.r : l < b.l;
  }
}q[maxn];

#define lson u << 1, l, mid
#define rson u << 1 | 1, mid + 1, r

void update(int u, int l, int r, int x, int v, int op){
  mi[u][op] = min(mi[u][op], v);
  if(l == r) return ;
  int mid = l + r >> 1;
  if(x <= mid) update(lson, x, v, op);
  else update(rson, x, v, op);
}

int query(int u, int l, int r, int ql, int qr, int v, int op){
  if(ql <= l && r <= qr) return mi[u][op] + v;
  int mid = l + r >> 1, res = inf;
  if(ql <= mid) res = min(res, query(lson, ql, qr, v, op));
  if(qr > mid) res = min(res, query(rson, ql, qr, v, op));
  return res;
}

void sol(){
  int ans = s[0][1];
  update(1, 0, n, 0, s[1][1], 1);
  update(1, 0, n, 0, s[0][1], 0);
  rep(i, 1, m){
	int l = q[i].l, r = q[i].r;
	int t = query(1, 0, n, l, r, -s[1][r + 1], 1);
	t = min(t, query(1, 0, n, 0, l - 1, -s[0][l] + s[1][l] - s[1][r + 1], 0));
	update(1, 0, n, r, t + s[1][r + 1], 1);
	update(1, 0, n, r, t + s[0][r + 1], 0);
	ans = min(ans, t + s[0][r + 1]);
  }
  printf("%d\n", ans);
}

int main(){
  memset(mi, 0x3f, sizeof(mi));
  rd(n);
  rep(i, 1, n){
	int x; rd(x);
	s[1 - x][i] = 1;
  }
  per(i, n, 1) rep(j, 0, 1) s[j][i] += s[j][i + 1];
  rd(m);
  rep(i, 1, m) rd(q[i].l), rd(q[i].r);
  sort(q + 1, q + m + 1);
  sol();
  return 0;
}

```