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;
int m;
int number;
set<int> dig0[17], dig1[17];
long long int sum;
set<int>::iterator iter;
int changeSum(int pos, int val) {
  long long int a = 1;
  int d = 0;
  while (true) {
    if (d == 17) return 0;
    if (a & val) {
      iter = dig1[d].lower_bound(pos);
      if (*iter != pos) {
        set<int>::iterator it = dig0[d].lower_bound(pos);
        iter = it;
        long long int r = *++it;
        long long int p = pos;
        long long int l = *--iter;
        sum += a * (r - l - 1) * (r - l) / 2 - a * (p - l - 1) * (p - l) / 2 -
               a * (r - p - 1) * (r - p) / 2;
        dig0[d].erase(pos);
        dig1[d].insert(pos);
      }
    } else {
      iter = dig0[d].lower_bound(pos);
      if (*iter != pos) {
        long long int p = pos;
        long long int r = *iter;
        long long int l = *--iter;
        sum -= a * (r - l - 1) * (r - l) / 2 - a * (r - p - 1) * (r - p) / 2 -
               a * (p - l - 1) * (p - l) / 2;
        dig1[d].erase(pos);
        dig0[d].insert(pos);
      }
    }
    ++d;
    a <<= 1;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < 17; i++) {
    dig0[i].insert(n);
    dig1[i].insert(n);
    dig0[i].insert(-1);
    dig1[i].insert(-1);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &number);
    int a = 1;
    int d = 0;
    while (true) {
      if (d == 17) break;
      if (a & number)
        dig1[d].insert(i);
      else
        dig0[d].insert(i);
      a <<= 1;
      ++d;
    }
  }
  long long int val = 1;
  for (int i = 0; i < 17; i++) {
    int p0 = 0;
    int pL = -1;
    int pR = -1;
    while (true) {
      p0 = *dig1[i].lower_bound(p0);
      if (p0 == n) break;
      iter = dig0[i].upper_bound(p0);
      pL = *iter;
      pR = (*--iter);
      long long int l = pL;
      long long int r = pR;
      sum += val * (l - r - 1) * (l - r) / 2;
      p0 = *dig0[i].lower_bound(p0);
    }
    val *= 2;
  }
  for (int i = 0; i < m; i++) {
    int pos, v;
    scanf("%d %d", &pos, &v);
    changeSum(pos - 1, v);
    printf("%I64d\n", sum);
  }
  return 0;
}