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;
const int N = 111111;
const int M = 17;
long long ans;
int a[N];
int n, m;
long long calc(long long x) { return (1 + x) * x / 2; }
struct BIT {
  int c[N];
  void clear() { memset(c, 0, sizeof(c)); }
  int lowbit(int x) { return x & (-x); }
  int add(int x, int v) {
    for (int i = x; i <= n; i += lowbit(i)) c[i] += v;
  }
  int sum(int x) {
    int res = 0;
    for (int i = x; i > 0; i -= lowbit(i)) res += c[i];
    return res;
  }
  int find_left(int x) {
    int t = sum(x);
    int l = 1, r = x;
    int res = -1;
    while (l <= r) {
      int mid = l + r >> 1;
      if (t - sum(mid - 1) == x - mid + 1) {
        res = mid, r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    return res;
  }
  int find_right(int x) {
    int t = sum(x - 1);
    int l = x, r = n;
    int res = -1;
    while (l <= r) {
      int mid = l + r >> 1;
      if (sum(mid) - t == mid - x + 1) {
        res = mid, l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    return res;
  }
} bit[M];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[n + 1] = 0;
  ans = 0;
  for (int i = 0; i < M; i++) {
    int cnt = 0;
    for (int j = 1; j <= n + 1; j++) {
      if (a[j] >> i & 1) {
        cnt++;
        bit[i].add(j, 1);
      } else {
        if (cnt) ans += calc(cnt) << i;
        cnt = 0;
      }
    }
  }
  int p, v;
  while (m--) {
    scanf("%d%d", &p, &v);
    int left, right;
    for (int i = 0; i < M; i++) {
      if ((a[p] ^ v) >> i & 1) {
        if (v >> i & 1) {
          if (p > 1 && (a[p - 1] >> i & 1)) {
            left = bit[i].find_left(p - 1);
            ans -= calc(p - left) << i;
          } else {
            left = p;
          }
          if (p < n && (a[p + 1] >> i & 1)) {
            right = bit[i].find_right(p + 1);
            ans -= calc(right - p) << i;
          } else {
            right = p;
          }
          bit[i].add(p, 1);
          ans += calc(right - left + 1) << i;
        } else {
          left = bit[i].find_left(p);
          right = bit[i].find_right(p);
          ans -= calc(right - left + 1) << i;
          ans += calc(p - left) << i;
          ans += calc(right - p) << i;
          bit[i].add(p, -1);
        }
        a[p] ^= 1 << i;
      }
    }
    cout << ans << endl;
  }
  return 0;
}