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;
int N, M, H;
int A[150010], B[150010];
struct segment_tree {
  int minv[150010 * 4], cov[150010 * 4];
  void push(int l, int r, int id) {
    cov[(id)*2 + 1] += cov[id];
    cov[(id)*2 + 2] += cov[id];
    minv[(id)*2 + 1] += cov[id];
    minv[(id)*2 + 2] += cov[id];
    cov[id] = 0;
  }
  void init(int l, int r, int id) {
    if (l == r) {
      minv[id] = M - l;
      return;
    }
    init(l, (l + r) / 2, (id)*2 + 1);
    init((l + r) / 2 + 1, r, (id)*2 + 2);
    minv[id] = min(minv[(id)*2 + 1], minv[(id)*2 + 2]);
  }
  void insert(int l, int r, int id, int a, int b, int v) {
    if (r < a || b < l) return;
    if (a <= l && r <= b) {
      cov[id] += v;
      minv[id] += v;
      return;
    }
    push(l, r, id);
    insert(l, (l + r) / 2, (id)*2 + 1, a, b, v);
    insert((l + r) / 2 + 1, r, (id)*2 + 2, a, b, v);
    minv[id] = min(minv[(id)*2 + 1], minv[(id)*2 + 2]);
  }
  int getmin() { return minv[0]; }
} seg;
int main() {
  scanf("%d%d%d", &N, &M, &H);
  for (int i = 0; i < M; i++) {
    scanf("%d", &B[i]);
  }
  sort(B, B + M);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    A[i] = lower_bound(B, B + M, H - A[i]) - B;
  }
  seg.init(0, M, 0);
  for (int i = 0; i < M - 1; i++) {
    seg.insert(0, M, 0, 0, A[i], -1);
  }
  int ans = 0;
  for (int i = M - 1; i < N; i++) {
    seg.insert(0, M, 0, 0, A[i], -1);
    int val = seg.getmin();
    if (val >= 0) {
      ans++;
    }
    seg.insert(0, M, 0, 0, A[i - M + 1], 1);
  }
  printf("%d\n", ans);
}
