Manao is solving a problem with the following statement:

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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;
const int N = 300000 + 10;
int sum[N << 2], mm[N << 2];
int vis[N];
void pushup(int o) {
  sum[o] = sum[o << 1] + sum[o << 1 | 1];
  mm[o] = min(mm[o << 1], sum[o << 1] + mm[o << 1 | 1]);
}
void build(int l, int r, int o) {
  if (l == r) {
    sum[o] = mm[o] = vis[l];
    return;
  }
  build(l, (l + r) / 2, o << 1);
  build((l + r) / 2 + 1, r, o << 1 | 1);
  pushup(o);
}
void update(int l, int r, int o, int x, int v) {
  if (l == r) {
    sum[o] = mm[o] = mm[o] + v;
    return;
  }
  if (x <= (l + r) / 2)
    update(l, (l + r) / 2, o << 1, x, v);
  else
    update((l + r) / 2 + 1, r, o << 1 | 1, x, v);
  pushup(o);
}
int sta[N], top;
int aa[N], bb[N];
int main() {
  int n, m, h;
  while (scanf("%d%d%d", &n, &m, &h) != EOF) {
    top = 1;
    for (int i = 1; i <= m; i++)
      scanf("%d", &bb[i]), bb[i] = h - bb[i], sta[top++] = bb[i];
    for (int i = 1; i <= n; i++) scanf("%d", &aa[i]), sta[top++] = aa[i];
    sort(sta + 1, sta + top);
    top = unique(sta + 1, sta + top) - sta;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= m; i++) {
      int id = lower_bound(sta + 1, sta + top, bb[i]) - sta;
      vis[id] += 1;
    }
    for (int i = 1; i <= m; i++) {
      int id = lower_bound(sta + 1, sta + top, aa[i]) - sta;
      vis[id] += -1;
    }
    build(1, top, 1);
    int cnt = 0;
    if (mm[1] >= 0) cnt++;
    for (int i = m + 1; i <= n; i++) {
      int id = lower_bound(sta + 1, sta + top, aa[i]) - sta;
      update(1, top, 1, id, -1);
      id = lower_bound(sta + 1, sta + top, aa[i - m]) - sta;
      update(1, top, 1, id, 1);
      if (mm[1] >= 0) cnt++;
    }
    printf("%d\n", cnt);
  }
  return 0;
}
