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 maxn = 150005;
int a[maxn];
int b[maxn];
int mMin[4 * maxn], mMax[4 * maxn], delta[4 * maxn];
void build(int v, int l, int r) {
  if (l == r) {
    mMin[v] = mMax[v] = l;
    return;
  }
  int c = (l + r) / 2;
  build(2 * v, l, c);
  build(2 * v + 1, c + 1, r);
  mMin[v] = min(mMin[2 * v], mMin[2 * v + 1]);
  mMax[v] = max(mMax[2 * v], mMax[2 * v + 1]);
}
void modify(int v, int l, int r, int i, int j, int d) {
  if (i <= l && r <= j) {
    mMin[v] += d;
    mMax[v] += d;
    delta[v] += d;
    return;
  }
  int c = (l + r) / 2;
  if (i <= c) modify(2 * v, l, c, i, j, d);
  if (c < j) modify(2 * v + 1, c + 1, r, i, j, d);
  mMin[v] = min(mMin[2 * v], mMin[2 * v + 1]) + delta[v];
  mMax[v] = max(mMax[2 * v], mMax[2 * v + 1]) + delta[v];
}
int resMin, resMax;
void get(int v, int l, int r, int i, int j, int d) {
  if (i <= l && r <= j) {
    resMin = min(resMin, mMin[v] + d);
    resMax = max(resMax, mMax[v] + d);
    return;
  }
  int c = (l + r) / 2;
  if (i <= c) get(2 * v, l, c, i, j, d + delta[v]);
  if (c < j) get(2 * v + 1, c + 1, r, i, j, d + delta[v]);
}
int was[maxn];
const int inf = 1000000005;
int main() {
  int n, len, h;
  scanf("%d %d %d", &n, &len, &h);
  for (int i = 0; i < len; i++) scanf("%d", b + i);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  sort(b, b + len);
  build(1, 1, len + 1);
  int ans = 0;
  int badCount = 0;
  for (int i = 0; i < n; i++) {
    if (i - len >= 0) {
      if (was[i - len] == len)
        badCount--;
      else
        modify(1, 1, len + 1, was[i - len] + 1, len + 1, +1);
    }
    was[i] = lower_bound(b, b + len, h - a[i]) - b;
    if (was[i] == len) {
      badCount++;
      continue;
    }
    modify(1, 1, len + 1, was[i] + 1, len + 1, -1);
    if (!badCount && i + 1 >= len) {
      resMax = -inf;
      resMin = inf;
      get(1, 1, len + 1, 1, len, 0);
      if (resMax <= 0) ans++;
    }
  }
  printf("%d", ans);
  return 0;
}
