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 = 150010, INF = ~0U >> 2;
struct node {
  int minv, mr;
} T[MAXN << 1];
int n, m, V0, A[MAXN], B[MAXN], C[MAXN], C2[MAXN], l0, _v, res = 0;
void init() {
  scanf("%d%d%d", &n, &m, &V0);
  for (int i = 0; i < m; i++) scanf("%d", &B[i]);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
}
void prepare() {
  sort(B, B + m);
  for (int i = 0; i < m; i++) B[i] = V0 - B[i];
  int l, r, mid, x;
  for (int i = 0; i < n; i++) {
    x = A[i];
    if (x < B[m - 1]) {
      C[i] = m;
      continue;
    } else {
      l = 0;
      r = m - 1;
    }
    while (l < r) {
      mid = l + r >> 1;
      if (x >= B[mid])
        r = mid;
      else
        l = mid + 1;
    }
    C[i] = l;
  }
  for (int i = 0; i < m; i++)
    if (C[i] < m) C2[C[i]]++;
  for (int i = 1; i < m; i++) C2[i] += C2[i - 1];
}
void mkt(int l, int r, int No) {
  if (l == r) {
    T[No].minv = C2[l] - l;
    T[No].mr = 0;
  } else {
    int mid = l + r >> 1, lch = (No + 1),
        rch = (No + r - l + 2 - ((r ^ l) & 1));
    mkt(l, mid, lch);
    mkt(mid + 1, r, rch);
    T[No].minv = T[lch].minv <= T[rch].minv ? T[lch].minv : T[rch].minv;
    T[No].mr = 0;
  }
}
void opr0(int l, int r, int No) {
  if (l >= l0) {
    T[No].minv += _v;
    T[No].mr += _v;
  } else {
    int mid = l + r >> 1, lch = (No + 1),
        rch = (No + r - l + 2 - ((r ^ l) & 1)), _;
    if (_ = T[No].mr) {
      T[No].mr = 0;
      T[lch].minv += _;
      T[lch].mr += _;
      T[rch].minv += _;
      T[rch].mr += _;
    }
    if (mid >= l0) opr0(l, mid, lch);
    opr0(mid + 1, r, rch);
    T[No].minv = T[lch].minv <= T[rch].minv ? T[lch].minv : T[rch].minv;
  }
}
void solve() {
  mkt(0, m - 1, 1);
  if (T[1].minv > 0) res++;
  for (int i = m; i < n; i++) {
    l0 = C[i - m];
    if (l0 < m) {
      _v = -1;
      opr0(0, m - 1, 1);
    }
    l0 = C[i];
    if (l0 < m) {
      _v = 1;
      opr0(0, m - 1, 1);
    }
    if (T[1].minv > 0) res++;
  }
}
void pri() { printf("%d\n", res); }
int main() {
  init();
  prepare();
  solve();
  pri();
  return 0;
}
