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 = 150000;
const int Inf = 2100000000;
struct Btree {
  int left, right;
  int tag;
  int min;
};
Btree tree[Maxn * 4];
inline void update(int x) {
  if (tree[x].left == tree[x].right) return;
  tree[x].min = min(tree[x * 2].min, tree[x * 2 + 1].min);
}
inline void plusNode(int x, int val) {
  tree[x].min += val;
  tree[x].tag += val;
}
inline void clean(int x) {
  if (tree[x].left == tree[x].right) return;
  if (tree[x].tag == 0) return;
  int tag = tree[x].tag;
  tree[x].tag = 0;
  plusNode(x * 2, tag);
  plusNode(x * 2 + 1, tag);
}
void build(int x, int left, int right) {
  tree[x].left = left;
  tree[x].right = right;
  tree[x].min = 0;
  if (left == right) {
    tree[x].min = -left;
  } else {
    int mid = (left + right) / 2;
    build(x * 2, left, mid);
    build(x * 2 + 1, mid + 1, right);
    update(x);
  }
}
void modify(int x, int left, int right, int val) {
  clean(x);
  if (left <= tree[x].left && tree[x].right <= right) {
    plusNode(x, val);
  } else {
    int mid = (tree[x].left + tree[x].right) / 2;
    if (left <= mid) modify(x * 2, left, right, val);
    if (mid < right) modify(x * 2 + 1, left, right, val);
    update(x);
  }
}
int query(int x, int left, int right) {
  clean(x);
  if (left <= tree[x].left && tree[x].right <= right) {
    return tree[x].min;
  } else {
    int ret = Inf;
    int mid = (tree[x].left + tree[x].right) / 2;
    if (left <= mid) ret = min(ret, query(x * 2, left, right));
    if (mid < right) ret = min(ret, query(x * 2 + 1, left, right));
    return ret;
  }
}
int A[Maxn], B[Maxn];
int n, len, H;
inline int find(int val) {
  int minNeed = H - val - 1;
  return upper_bound(B + 1, B + len + 1, minNeed) - (B);
}
int Ans = 0;
int pos;
int main() {
  scanf("%d%d%d", &n, &len, &H);
  for (int i = 1; i <= len; i++) scanf("%d", &B[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  sort(B + 1, B + len + 1);
  build(1, 1, n);
  for (int i = 1; i <= len; i++) {
    pos = find(A[i]);
    if (pos <= len) modify(1, pos, len, 1);
  }
  if (query(1, 1, len) >= 0) Ans++;
  for (int i = len + 1; i <= n; i++) {
    pos = find(A[i - len]);
    if (pos <= len) modify(1, pos, len, -1);
    pos = find(A[i]);
    if (pos <= len) modify(1, pos, len, 1);
    if (query(1, 1, len) >= 0) Ans++;
  }
  printf("%d\n", Ans);
  return 0;
}
