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;
long long calc(long long x) { return x * (x + 1) / 2; }
struct SegTree {
  struct Node {
    int lm, rm;
    long long sum;
  } q[100010 << 2];
  void pushUp(int x, int l, int r, int m) {
    q[x].lm = q[((x) << 1)].lm;
    if (q[((x) << 1)].lm == m - l + 1) q[x].lm += q[(((x) << 1) | 1)].lm;
    q[x].rm = q[(((x) << 1) | 1)].rm;
    if (q[(((x) << 1) | 1)].rm == r - m) q[x].rm += q[((x) << 1)].rm;
    q[x].sum = q[((x) << 1)].sum - calc(q[((x) << 1)].rm) +
               q[(((x) << 1) | 1)].sum - calc(q[(((x) << 1) | 1)].lm) +
               calc(q[((x) << 1)].rm + q[(((x) << 1) | 1)].lm);
  }
  void build(int x, int l, int r, const int *a) {
    int mid = (((l) & (r)) + (((l) ^ (r)) >> 1));
    if (l == r) {
      q[x].lm = q[x].rm = a[l];
      q[x].sum = a[l];
      return;
    }
    build(((x) << 1), l, mid, a);
    build((((x) << 1) | 1), mid + 1, r, a);
    pushUp(x, l, r, mid);
  }
  void update(int x, int l, int r, int p, int val) {
    int mid = (((l) & (r)) + (((l) ^ (r)) >> 1));
    if (l == r) {
      q[x].lm = q[x].rm = val;
      q[x].sum = val;
      return;
    }
    if (p <= mid)
      update(((x) << 1), l, mid, p, val);
    else
      update((((x) << 1) | 1), mid + 1, r, p, val);
    pushUp(x, l, r, mid);
  }
  long long query(int x, int l, int r, int low, int high) {
    int mid = (((l) & (r)) + (((l) ^ (r)) >> 1));
    long long ret = 0;
    if (low <= l && r <= high) return q[x].sum;
    if (low <= mid) ret += query(((x) << 1), l, mid, low, high);
    if (high > mid) ret += query((((x) << 1) | 1), mid + 1, r, low, high);
    return ret;
  }
} st[21];
int a[21][100010], t[100010];
long long getSum() {
  int i;
  long long ret = 0;
  for (i = 0; i < 20; ++i) {
    ret += (st[i].q[1].sum << i);
  }
  return ret;
}
int main() {
  int n, m, i, j, p, x;
  long long ans;
  while (~scanf("%d %d", &n, &m)) {
    for (i = 0; i < n; ++i) {
      scanf("%d", &t[i]);
      for (j = 0; j < 20; ++j) a[j][i] = ((t[i] >> j) & 1);
    }
    for (i = 0; i < 20; ++i) st[i].build(1, 0, n - 1, a[i]);
    for (i = 0; i < m; ++i) {
      scanf("%d %d", &p, &x);
      t[p - 1] = x;
      for (j = 0; j < 20; ++j) st[j].update(1, 0, n - 1, p - 1, (x >> j) & 1);
      printf("%I64d\n", getSum());
    }
  }
  return 0;
}