Manao is solving a problem with the following statement:

<image>

He came up with a solution that produces the correct answers but is too slow. You are given the pseudocode of his solution, where the function getAnswer calculates the answer to the problem:
    
    
      
    getAnswer(a[1..n], b[1..len], h)  
      answer = 0  
      for i = 1 to n-len+1  
        answer = answer + f(a[i..i+len-1], b, h, 1)  
      return answer  
      
    f(s[1..len], b[1..len], h, index)  
      if index = len+1 then  
        return 1  
      for i = 1 to len  
        if s[index] + b[i] >= h  
          mem = b[i]  
          b[i] = 0  
          res = f(s, b, h, index + 1)  
          b[i] = mem  
          if res > 0  
            return 1  
      return 0  
    

Your task is to help Manao optimize his algorithm.

Input

The first line contains space-separated integers n, len and h (1 ≤ len ≤ n ≤ 150000; 1 ≤ h ≤ 109). The second line contains len space-separated integers b1, b2, ..., blen (1 ≤ bi ≤ 109). The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single number — the answer to Manao's problem.

Examples

Input

5 2 10
5 3
1 8 5 5 7


Output

2
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
namespace Mymath {
long long qp(long long x, long long p, long long mod) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * x % mod;
    x = x * x % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long x, long long mod) { return qp(x, mod - 2, mod); }
long long C(long long N, long long K, long long fact[], long long mod) {
  return fact[N] * inv(fact[K], mod) % mod * inv(fact[N - K], mod) % mod;
}
template <typename Tp>
Tp gcd(Tp A, Tp B) {
  if (B == 0) return A;
  return gcd(B, A % B);
}
template <typename Tp>
Tp lcm(Tp A, Tp B) {
  return A * B / gcd(A, B);
}
};  // namespace Mymath
namespace fwt {
using namespace Mymath;
void FWT(int a[], int n, long long mod) {
  for (int d = 1; d < n; d <<= 1)
    for (int m = d << 1, i = 0; i < n; i += m)
      for (int j = 0; j < d; j++) {
        int x = a[i + j], y = a[i + j + d];
        a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
      }
}
void UFWT(int a[], int n, long long mod) {
  long long rev = inv(2, mod);
  for (int d = 1; d < n; d <<= 1)
    for (int m = d << 1, i = 0; i < n; i += m)
      for (int j = 0; j < d; j++) {
        int x = a[i + j], y = a[i + j + d];
        a[i + j] = 1LL * (x + y) * rev % mod,
              a[i + j + d] = (1LL * (x - y) * rev % mod + mod) % mod;
      }
}
void solve(int a[], int b[], int n, long long mod) {
  FWT(a, n, mod);
  FWT(b, n, mod);
  for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * b[i] % mod;
  UFWT(a, n, mod);
}
};  // namespace fwt
namespace Maxflow {
const int Maxn = 1005;
struct edge {
  int to, rev, cap;
  edge(int _to, int _rev, int _cap) { to = _to, rev = _rev, cap = _cap; }
};
int src = 1002, snk = 1003;
vector<edge> G[Maxn];
int dist[Maxn], seen[Maxn];
int n;
void Clear() {
  for (int i = 0; i < Maxn; i++) G[i].clear();
}
void bfs(int s) {
  for (int i = 0; i < Maxn; i++) dist[i] = -1;
  dist[s] = 0;
  queue<int> Q;
  Q.push(s);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < G[x].size(); i++) {
      edge &e = G[x][i];
      if (e.cap && dist[e.to] == -1) {
        dist[e.to] = dist[x] + 1;
        Q.push(e.to);
      }
    }
  }
}
int dfs(int now, int dest, int fl) {
  if (now == dest) return fl;
  for (int &i = seen[now]; i < G[now].size(); i++) {
    edge &ed = G[now][i];
    if (ed.cap && dist[ed.to] == dist[now] + 1) {
      int f = dfs(ed.to, dest, min(fl, ed.cap));
      if (f) {
        ed.cap -= f;
        G[ed.to][ed.rev].cap += f;
        return f;
      }
    }
  }
  return 0;
}
int dinic(int s = src, int t = snk) {
  int ret = 0;
  while (1) {
    memset(seen, 0, sizeof(seen));
    bfs(s);
    if (dist[t] == -1) return ret;
    int f = dfs(s, t, 2147483647);
    while (f) {
      ret += f;
      f = dfs(s, t, 2147483647);
    }
  }
}
};  // namespace Maxflow
namespace Geometry {
struct iP {
  long long x, y;
  iP(long long _x, long long _y) { x = _x, y = _y; }
  iP() { x = y = 0; }
  iP operator+(iP p) { return iP(x + p.x, y + p.y); }
  iP operator-(iP p) { return iP(x - p.x, y - p.y); }
  iP operator*(long long k) { return iP(x * k, y * k); }
  long long det(iP p) { return x * p.y - y * p.x; }
  long long dot(iP p) { return x * p.x + y * p.y; }
  long long dist() { return x * x + y * y; }
  bool operator<(const iP &p) const {
    if (x != p.x) return x < p.x;
    return y < p.y;
  }
};
struct dP {
  double x, y;
  dP(double _x, double _y) { x = _x, y = _y; }
  dP() { x = y = 0.0; }
  dP operator+(dP p) { return dP(x + p.x, y + p.y); }
  dP operator-(dP p) { return dP(x - p.x, y - p.y); }
  dP operator*(double k) { return dP(x * k, y * k); }
  double det(dP p) { return x * p.y - y * p.x; }
  double dot(dP p) { return x * p.x + y * p.y; }
  double dist() { return x * x + y * y; }
  bool operator<(const dP &p) const {
    if (x != p.x) return x < p.x;
    return y < p.y;
  }
};
vector<iP> convex_hull(iP X[], int l) {
  vector<iP> ret;
  sort(X + 1, X + 1 + l);
  ret.push_back(X[1]);
  for (int i = 2; i <= l; i++) {
    while (ret.size() >= 2 &&
           (ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
      ret.pop_back();
    }
    ret.push_back(X[i]);
  }
  int rs = ret.size() + 1;
  for (int i = l - 1; i >= 1; i--) {
    while (ret.size() >= rs &&
           (ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
      ret.pop_back();
    }
    ret.push_back(X[i]);
  }
  return ret;
}
};  // namespace Geometry
const int Maxn = 150005;
int n, len, h;
int a[Maxn], b[Maxn];
int p[Maxn];
int tree[Maxn * 4];
int tag[Maxn * 4];
void build(int p, int l, int r) {
  if (l == r) {
    tree[p] = -l;
    return;
  }
  int mid = l + r >> 1;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
  tree[p] = min(tree[p * 2], tree[p * 2 + 1]);
}
void Push(int p) {
  if (tag[p]) {
    tag[p * 2] += tag[p];
    tag[p * 2 + 1] += tag[p];
    tree[p * 2] += tag[p];
    tree[p * 2 + 1] += tag[p];
    tag[p] = 0;
  }
}
void modify(int p, int l, int r, int lo, int hi, int val) {
  if (lo > hi) return;
  if (lo <= l && r <= hi) {
    tree[p] += val;
    tag[p] += val;
    return;
  }
  int mid = l + r >> 1;
  Push(p);
  if (lo <= mid) {
    modify(p * 2, l, mid, lo, min(hi, mid), val);
  }
  if (hi > mid) {
    modify(p * 2 + 1, mid + 1, r, max(lo, mid + 1), hi, val);
  }
  tree[p] = min(tree[p * 2], tree[p * 2 + 1]);
}
int main() {
  n = read();
  len = read();
  h = read();
  for (int i = 1; i <= len; i++) b[i] = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(b + 1, b + 1 + len);
  for (int i = 1; i <= n; i++) {
    p[i] = lower_bound(b + 1, b + 1 + len, h - a[i]) - b;
  }
  build(1, 1, len);
  for (int i = 1; i <= len; i++) {
    modify(1, 1, len, p[i], len, 1);
  }
  int ans = 0;
  for (int i = len; i <= n; i++) {
    int v = tree[1];
    if (v >= 0) {
      ans++;
    }
    if (i == n) break;
    modify(1, 1, len, p[i + 1], len, 1);
    modify(1, 1, len, p[i - len + 1], len, -1);
  }
  printf("%d\n", ans);
  return ~~(0 ^ 0 ^ 0);
}
