Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
const int MAXQ = 200000;
const int MOD = 1000000007;
const int MAXS = 4 * MAXN;
void inc(int &a, int b) {
  if ((a += b) >= MOD) a -= MOD;
}
int n, nq;
int a[MAXN];
int w[MAXN];
int qa[MAXQ], qb[MAXQ], qans[MAXQ];
long long ssum[MAXS];
int ssum2[MAXS];
void spull(int x) {
  ssum[x] = ssum[2 * x + 1] + ssum[2 * x + 2];
  ssum2[x] = (ssum2[2 * x + 1] + ssum2[2 * x + 2]) % MOD;
}
void sinit(int x, int l, int r) {
  if (l == r) {
    int idx = l;
    ssum[x] = w[idx];
    ssum2[x] = (long long)(a[idx] - idx) * w[idx] % MOD;
  } else {
    int m = l + (r - l) / 2;
    sinit(2 * x + 1, l, m);
    sinit(2 * x + 2, m + 1, r);
    spull(x);
  }
}
void sset(int x, int l, int r, int IDX, int VAL) {
  if (l == r) {
    ssum[x] = VAL;
    ssum2[x] = (long long)(a[IDX] - IDX) * VAL % MOD;
  } else {
    int m = l + (r - l) / 2;
    if (IDX <= m) sset(2 * x + 1, l, m, IDX, VAL);
    if (m + 1 <= IDX) sset(2 * x + 2, m + 1, r, IDX, VAL);
    spull(x);
  }
}
pair<long long, int> sget(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    return make_pair(ssum[x], ssum2[x]);
  } else {
    int m = l + (r - l) / 2;
    pair<long long, int> ret = make_pair(0LL, 0);
    if (L <= m) {
      pair<long long, int> cur = sget(2 * x + 1, l, m, L, R);
      ret.first += cur.first;
      inc(ret.second, cur.second);
    }
    if (m + 1 <= R) {
      pair<long long, int> cur = sget(2 * x + 2, m + 1, r, L, R);
      ret.first += cur.first;
      inc(ret.second, cur.second);
    }
    return ret;
  }
}
int sfind(int x, int l, int r, int L, int R, long long &want) {
  if (l == r && want <= ssum[x]) {
    return l;
  } else if (L <= l && r <= R && ssum[x] < want) {
    want -= ssum[x];
    return -1;
  } else {
    int m = l + (r - l) / 2;
    if (L <= m) {
      int cur = sfind(2 * x + 1, l, m, L, R, want);
      if (cur != -1) return cur;
    }
    if (m + 1 <= R) {
      int cur = sfind(2 * x + 2, m + 1, r, L, R, want);
      if (cur != -1) return cur;
    }
    return -1;
  }
}
void solve() {
  sinit(0, 0, n - 1);
  for (int i = (0); i < (nq); ++i) {
    if (qa[i] < 0) {
      int idx = -qa[i] - 1, nw = qb[i];
      sset(0, 0, n - 1, idx, nw);
    } else {
      int l = qa[i] - 1, r = qb[i] - 1;
      long long wsum = sget(0, 0, n - 1, l, r).first;
      long long want = (wsum + 1) / 2;
      int mid = sfind(0, 0, n - 1, l, r, want);
      assert(mid != -1);
      pair<long long, int> lres = sget(0, 0, n - 1, l, mid);
      pair<long long, int> rres = sget(0, 0, n - 1, mid, r);
      qans[i] = (((long long)(a[mid] - mid) * (lres.first % MOD) - lres.second +
                  (mid - a[mid]) * (rres.first % MOD) + rres.second) %
                     MOD +
                 MOD) %
                MOD;
    }
  }
}
void run() {
  scanf("%d%d", &n, &nq);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  for (int i = (0); i < (n); ++i) scanf("%d", &w[i]);
  for (int i = (0); i < (nq); ++i) scanf("%d%d", &qa[i], &qb[i]);
  solve();
  for (int i = (0); i < (nq); ++i)
    if (qa[i] > 0) printf("%d\n", qans[i]);
}
int main() {
  run();
  return 0;
}