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 long long MAXN = 1e5 + 5;
const long long MAXBIT = 18;
long long N, M;
long long A[MAXN];
struct Node;
Node *newNode();
struct Node {
  Node *pl, *pr;
  long long nl = 0, nr = 0;
  long long lrun = 0, rrun = 0, val = 0;
  bool isFull = 0;
  void updateVal() {
    lrun = 0, rrun = 0, val = 0, isFull = false;
    val = pl->val + pr->val;
    if (pl->isFull && pr->isFull) {
      isFull = true;
      return;
    }
    isFull = false;
    if (pl->isFull) {
      lrun = (pl->nr - pl->nl) + pr->lrun;
      rrun = pr->rrun;
    } else if (pr->isFull) {
      lrun = pl->lrun;
      rrun = (pr->nr - pr->nl) + pl->rrun;
    } else {
      lrun = pl->lrun;
      rrun = pr->rrun;
      val += (pl->rrun + pr->lrun) * (pl->rrun + pr->lrun + 1) / 2;
    }
  }
  void build(long long l, long long r, long long b) {
    nl = l, nr = r;
    if (nl + 1 == nr) {
      isFull = (A[nl] & (1 << b)) != 0;
      return;
    }
    (pl = newNode())->build(nl, nl + nr >> 1, b);
    (pr = newNode())->build(nl + nr >> 1, nr, b);
    updateVal();
  }
  void modify(long long p, long long x) {
    if (p < nl || nr <= p) {
      return;
    }
    if (nl + 1 == nr) {
      isFull = x;
      return;
    }
    pl->modify(p, x);
    pr->modify(p, x);
    updateVal();
  }
};
long long bufSize = MAXN * 2 * MAXBIT;
Node buf[MAXN * 2 * MAXBIT];
Node *newNode() { return &buf[--bufSize]; }
Node *seg[MAXBIT];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (long long i = 0; i < N; i++) {
    cin >> A[i];
  }
  for (long long b = 0; b < MAXBIT; b++) {
    seg[b] = newNode();
    seg[b]->build(0, N, b);
  }
  for (long long i = 0; i < M; i++) {
    long long p, v;
    cin >> p >> v;
    p--;
    for (long long b = 0; b < MAXBIT; b++) {
      seg[b]->modify(p, (v & (1 << b)) != 0);
    }
    long long ans = 0;
    for (long long b = 0; b < MAXBIT; b++) {
      long long cur = (seg[b]->val + (seg[b]->lrun * (seg[b]->lrun + 1)) / 2 +
                       (seg[b]->rrun * (seg[b]->rrun + 1)) / 2);
      cur += (seg[b]->isFull) * (seg[b]->nr - seg[b]->nl) *
             (seg[b]->nr - seg[b]->nl + 1) / 2;
      ans += cur * (1 << b);
    }
    cout << ans << endl;
  }
}