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 n, m, a[100005], x, y;
long long sum;
void init() {
  for (int i = 0; (1 << i) < 100005; i++) {
    int k = 1 << i;
    long long tot = 0;
    for (int j = 1; j <= n; j++) {
      if (k & a[j])
        tot++;
      else {
        sum += tot * (tot + 1) / 2 * (long long)k;
        tot = 0;
      }
    }
    if (tot) sum += tot * (tot + 1) / 2 * (long long)k;
  }
}
void work(int x, int y) {
  for (int i = 0; (1 << i) < 100005; i++) {
    int k = 1 << i;
    if ((k & y) == (k & a[x])) continue;
    long long l = 0, r = 0;
    for (int p = x - 1; p > 0; p--)
      if (k & a[p])
        l++;
      else
        break;
    for (int p = x + 1; p <= n; p++)
      if (k & a[p])
        r++;
      else
        break;
    long long tmp = (l * r + l + r + 1) * (long long)k;
    if (k & y)
      sum += tmp;
    else
      sum -= tmp;
  }
}
int main() {
  cin >> n >> m;
  sum = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  init();
  while (m--) {
    cin >> x >> y;
    work(x, y);
    cout << sum << endl;
    a[x] = y;
  }
}