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;
bool a[25][100001];
int main() {
  int n, m;
  cin >> n >> m;
  int temp;
  for (int i = 1; i < n + 1; i++) {
    cin >> temp;
    for (int j = 0; j < 25; j++) {
      a[j][i] = ((temp & (1 << j)) > 0);
    }
  }
  long long int count = 0;
  for (int i = 0; i < 25; i++) {
    long long int bcount = 0;
    for (int j = 1; j < n + 1; j++) {
      if (a[i][j])
        bcount++;
      else {
        count += ((long long)bcount * (bcount + 1) / 2) << i;
        bcount = 0;
      }
    }
    count += (1 << i) * bcount * (bcount + 1) / 2;
  }
  int p, k;
  for (int i = 0; i < m; i++) {
    cin >> p >> k;
    for (int j = 0; j < 25; j++) {
      if (((k & (1 << j)) > 0) == a[j][p]) continue;
      int i_l, i_h;
      for (i_l = p - 1; i_l > 0 && a[j][i_l]; i_l--)
        ;
      for (i_h = p + 1; i_h <= n && a[j][i_h]; i_h++)
        ;
      if ((k & (1 << j)) > 0) {
        a[j][p] = 1;
        count = count + (((long long)(p - i_l) * (i_h - p)) << j);
      } else if ((k & (1 << j)) == 0) {
        a[j][p] = 0;
        count = count - (((long long)(p - i_l) * (i_h - p)) << j);
      }
    }
    cout << count << endl;
  }
  return (0);
}