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, m, a[112345];
map<int, int> s[20];
long long gao(int c, int i) { return c * (long long)(c + 1) / 2 * (1L << i); }
int get(int x, int i) { return x & (1 << i); }
void out(map<int, int> &s) {
  cout << "map:" << endl;
  for (map<int, int>::iterator it = s.begin(); it != s.end(); it++)
    cout << it->first << ' ' << it->second << endl;
  cout << endl;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (int)n; i++) scanf("%d", a + i);
  long long sum = 0;
  for (int j = 0; j < (int)20; j++) {
    a[n] = 0;
    int c = 0, p = -1;
    for (int i = 0; i <= n; i++) {
      if (a[i] & (1 << j)) {
        if (!c++) p = i;
      } else {
        if (c) {
          s[j][p] = c;
          sum += gao(c, j);
        }
        c = 0;
      }
    }
  }
  while (m--) {
    int p, v;
    scanf("%d%d", &p, &v);
    p--;
    for (int i = 0; i < (int)20; i++) {
      map<int, int> &ss = s[i];
      if (ss.size()) {
      }
      if (get(v, i) != get(a[p], i)) {
        if (get(v, i)) {
          if (p + 1 < n && get(a[p + 1], i)) {
            int t;
            sum += gao(t = ss[p] = ss[p + 1] + 1, i);
            sum -= gao(t - 1, i);
            ss.erase(p + 1);
          } else {
            ss[p] = 1;
            sum += gao(1, i);
          }
          if (p - 1 >= 0 && get(a[p - 1], i)) {
            int t, t0;
            map<int, int>::iterator it;
            it = ss.lower_bound(p);
            it--;
            int x = it->first, y = it->second;
            sum += gao(ss[x] = y + ss[p], i);
            sum -= gao(ss[p], i);
            sum -= gao(y, i);
            ss.erase(p);
          }
        } else {
          map<int, int>::iterator it;
          it = ss.upper_bound(p);
          it--;
          int x = it->first, y = it->second;
          assert(x <= p && x + y > p);
          sum -= gao(y, i);
          ss.erase(x);
          if (x < p) {
            ss[x] = p - x;
            sum += gao(p - x, i);
          }
          if (x + y - 1 > p) {
            ss[p + 1] = x + y - 1 - p;
            sum += gao(x + y - 1 - p, i);
          }
        }
      }
    }
    a[p] = v;
    cout << sum << "\n";
  }
}