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;
const int maxn = 1e5 + 5;
int a[maxn], n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  long long sum = 0;
  for (int i = 0; i < 18; i++) {
    int k = 1 << i;
    int cnt = 0;
    for (int j = 1; j <= n; j++) {
      if (k & a[j])
        cnt++;
      else {
        sum += 1LL * cnt * (cnt + 1) / 2 * k;
        cnt = 0;
      }
    }
    if (cnt) sum += 1LL * cnt * (cnt + 1) / 2 * k;
  }
  while (m--) {
    int p, v;
    scanf("%d%d", &p, &v);
    for (int i = 0; i < 18; i++) {
      int k = 1 << i;
      if ((k & v) == (k & a[p])) continue;
      int l = 0, r = 0;
      for (int j = p - 1; j >= 1; j--) {
        if (a[j] & k)
          l++;
        else
          break;
      }
      for (int j = p + 1; j <= n; j++) {
        if (a[j] & k)
          r++;
        else
          break;
      }
      long long tmp = (l + r + 1LL * l * r + 1) * k;
      if (k & v)
        sum += tmp;
      else
        sum -= tmp;
    }
    a[p] = v;
    printf("%lld\n", sum);
  }
  return 0;
}