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;
long long MOD = 1000000007;
long long dr[800000];
long long pr[800000];
long long sf[800000];
long long w[800000];
long long a[800000];
void bl(long long v, long long l, long long r) {
  if (r == l + 1) {
    dr[v] = w[l];
    pr[v] = 0;
    sf[v] = 0;
    return;
  }
  bl(2 * v + 1, l, (l + r) / 2);
  bl(2 * v + 2, (l + r) / 2, r);
  dr[v] = dr[2 * v + 1] + dr[2 * v + 2];
  pr[v] = pr[2 * v + 1] + pr[2 * v + 2] +
          (a[r - 1] - (r - (l + r) / 2) - a[(r + l) / 2 - 1]) *
              (dr[2 * v + 1] % MOD);
  sf[v] = sf[2 * v + 1] + sf[2 * v + 2] +
          (a[(r + l) / 2] - a[l] - ((r + l) / 2 - l)) * (dr[2 * v + 2] % MOD);
  sf[v] %= MOD;
  pr[v] %= MOD;
}
void ch(long long v, long long l, long long r, long long id, long long x) {
  if (r == l + 1 && l == id) {
    dr[v] = x;
    pr[v] = 0;
    sf[v] = 0;
    return;
  }
  if (r <= id || l > id) {
    return;
  }
  ch(2 * v + 1, l, (l + r) / 2, id, x);
  ch(2 * v + 2, (l + r) / 2, r, id, x);
  dr[v] = dr[2 * v + 1] + dr[2 * v + 2];
  pr[v] = pr[2 * v + 1] + pr[2 * v + 2] +
          (a[r - 1] - (r - (l + r) / 2) - a[(r + l) / 2 - 1]) *
              (dr[2 * v + 1] % MOD);
  sf[v] = sf[2 * v + 1] + sf[2 * v + 2] +
          (a[(r + l) / 2] - a[l] - ((r + l) / 2 - l)) * (dr[2 * v + 2] % MOD);
  sf[v] %= MOD;
  pr[v] %= MOD;
}
long long sm(long long v, long long l, long long r, long long ln,
             long long rn) {
  if (ln >= r || rn <= l) {
    return 0;
  }
  if (ln <= l && r <= rn) {
    return dr[v];
  }
  return sm(2 * v + 1, l, (l + r) / 2, ln, rn) +
         sm(2 * v + 2, (l + r) / 2, r, ln, rn);
}
pair<long long, long long> indd(long long v, long long l, long long r,
                                long long k, long long ln, long long rn) {
  if (ln >= r || rn <= l) {
    return {0, -1};
  }
  if (l == r - 1) {
    return {dr[v], l};
  }
  if (ln <= l && r <= rn) {
    if (dr[v] < k) return {dr[v], r};
  }
  pair<long long, long long> cr =
      indd(2 * v + 1, l, (l + r) / 2, k, ln, min(rn, (l + r) / 2));
  if (cr.first < k) {
    pair<long long, long long> cr2 =
        indd(2 * v + 2, (l + r) / 2, r, k - cr.first, max(ln, (l + r) / 2), rn);
    return {cr2.first + cr.first, cr2.second};
  } else {
    return cr;
  }
}
long long prr(long long v, long long l, long long r, long long ln,
              long long k) {
  if (ln >= r || k <= l) {
    return 0;
  }
  if (ln <= l && r <= k) {
    return (pr[v] + (a[k - 1] - (k - r) - a[r - 1]) * (dr[v] % MOD)) % MOD;
  }
  return (prr(2 * v + 1, l, (l + r) / 2, ln, k) +
          prr(2 * v + 2, (l + r) / 2, r, ln, k)) %
         MOD;
}
long long sff(long long v, long long l, long long r, long long k,
              long long rn) {
  if (k >= r || rn <= l) {
    return 0;
  }
  if (k <= l && r <= rn) {
    return (sf[v] + (a[l] - a[k] - (l - k)) * (dr[v] % MOD)) % MOD;
  }
  return (sff(2 * v + 1, l, (l + r) / 2, k, rn) +
          sff(2 * v + 2, (l + r) / 2, r, k, rn)) %
         MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> w[i];
  }
  bl(0, 0, n);
  for (int i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      ch(0, 0, n, -x - 1, y);
    } else {
      x--;
      long long aaa = (sm(0, 0, n, x, y) + 1) / 2;
      long long L = indd(0, 0, n, aaa, x, y).second;
      cout << (prr(0, 0, n, x, L + 1) + sff(0, 0, n, L, y)) % MOD << endl;
    }
  }
}