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;
struct node {
  int left, right;
  long long ans, lmax, rmax;
  node *lc, *rc;
};
int get() {
  char c;
  while (c = getchar(), (c < '0' || c > '9') && (c != '-'))
    ;
  bool flag = (c == '-');
  if (flag) c = getchar();
  int x = 0;
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - 48;
    c = getchar();
  }
  return flag ? -x : x;
}
void output(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int len = 0, data[20];
  while (x) {
    data[len++] = x % 10;
    x /= 10;
  }
  if (!len) data[len++] = 0;
  while (len--) putchar(data[len] + 48);
  putchar('\n');
}
int data[100000];
void combine(node *p) {
  p->ans = p->lc->ans + p->rc->ans + p->lc->rmax * p->rc->lmax;
  p->lmax = p->lc->lmax;
  if (p->lmax == p->lc->right - p->lc->left + 1) p->lmax += p->rc->lmax;
  p->rmax = p->rc->rmax;
  if (p->rmax == p->rc->right - p->rc->left + 1) p->rmax += p->lc->rmax;
}
node *build(int l, int r) {
  node *p = new node;
  p->left = l;
  p->right = r;
  if (l == r)
    p->ans = p->lmax = p->rmax = data[l];
  else {
    int mid = (l + r) / 2;
    p->lc = build(l, mid);
    p->rc = build(mid + 1, r);
    combine(p);
  }
  return p;
}
void update(node *p, int x, int y) {
  if (p->left == p->right)
    p->ans = p->lmax = p->rmax = y;
  else {
    update((x <= (p->left + p->right) / 2) ? p->lc : p->rc, x, y);
    combine(p);
  }
}
int main() {
  int n = get(), m = get();
  static int a[100000];
  for (int i = 0; i < n; i++) a[i] = get();
  node *t[17];
  for (int i = 0; i < 17; i++) {
    for (int j = 0; j < n; j++) data[j] = (a[j] >> i) & 1;
    t[i] = build(0, n - 1);
  }
  while (m--) {
    int x = get() - 1, y = get();
    long long ans = 0;
    for (int i = 0; i < 17; i++) {
      update(t[i], x, (y >> i) & 1);
      ans += t[i]->ans << i;
    }
    output(ans);
  }
  return 0;
}