Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int A[100008];
int main() {
  long long ans = 0;
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  for (int bit = 0; bit <= 18; bit++) {
    long long idx = 0, tot = 0;
    while (idx < n) {
      if (A[idx] & (1 << bit))
        tot++;
      else
        ans += (1 << bit) * (tot * (tot + 1) / 2), tot = 0;
      idx++;
    }
    ans += (1 << bit) * (tot * (tot + 1) / 2);
  }
  int p, v;
  while (m--) {
    scanf("%d %d", &p, &v);
    p--;
    for (int bit = 0; bit <= 18; bit++) {
      long long left = 0, right = 0;
      for (int i = p - 1; i >= 0; i--)
        if (A[i] & (1 << bit))
          left++;
        else
          break;
      for (int i = p + 1; i < n; i++)
        if (A[i] & (1 << bit))
          right++;
        else
          break;
      if (A[p] & (1 << bit)) ans -= (left + 1) * (right + 1) * (1 << bit);
      if (v & (1 << bit)) ans += (left + 1) * (right + 1) * (1 << bit);
    }
    A[p] = v;
    printf("%lld\n", ans);
  }
  return 0;
}