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;
const int P = 1e9 + 7;
const int maxn = 2e5;
const int maxnode = maxn << 2;
int n, q;
long long p[maxn + 5], ai[maxn + 5], si[maxn + 5];
long long tr[maxnode + 5], lz[maxnode + 5];
void Add(int x, long long v) {
  for (; x <= n; x += x & (-x)) si[x] += v;
}
long long Sum(int x) {
  long long r = 0;
  for (; x; x -= x & (-x)) r += si[x];
  return r;
}
void Build(int x, int l, int r) {
  if (l == r)
    tr[x] = Sum(l) % P * (p[l + 1] - p[l] - 1) % P;
  else {
    Build((x << 1), l, ((l + r) >> 1));
    Build(((x << 1) | 1), ((l + r) >> 1) + 1, r);
    tr[x] = (tr[(x << 1)] + tr[((x << 1) | 1)]) % P;
  }
}
void PushDown(int x, int l, int r) {
  lz[(x << 1)] = (lz[(x << 1)] + lz[x]) % P;
  lz[((x << 1) | 1)] = (lz[((x << 1) | 1)] + lz[x]) % P;
  tr[(x << 1)] =
      (tr[(x << 1)] +
       (p[((l + r) >> 1) + 1] - p[l] - (((l + r) >> 1) + 1 - l)) * lz[x]) %
      P;
  tr[((x << 1) | 1)] =
      (tr[((x << 1) | 1)] +
       (p[r + 1] - p[((l + r) >> 1) + 1] - (r - ((l + r) >> 1))) * lz[x]) %
      P;
  lz[x] = 0;
}
void Upd(int x, int l, int r, int ll, int rr, long long v) {
  if (ll <= l && r <= rr) {
    lz[x] = (lz[x] + v) % P;
    tr[x] = (tr[x] + (p[r + 1] - p[l] - (r + 1 - l)) * v) % P;
  } else {
    PushDown(x, l, r);
    if (ll <= ((l + r) >> 1)) Upd((x << 1), l, ((l + r) >> 1), ll, rr, v);
    if (((l + r) >> 1) < rr)
      Upd(((x << 1) | 1), ((l + r) >> 1) + 1, r, ll, rr, v);
    tr[x] = (tr[(x << 1)] + tr[((x << 1) | 1)]) % P;
  }
}
long long Query(int x, int l, int r, int ll, int rr) {
  if (ll <= l && r <= rr) return tr[x];
  PushDown(x, l, r);
  long long ret = 0;
  if (ll <= ((l + r) >> 1)) ret = Query((x << 1), l, ((l + r) >> 1), ll, rr);
  if (((l + r) >> 1) < rr)
    ret = (ret + Query(((x << 1) | 1), ((l + r) >> 1) + 1, r, ll, rr)) % P;
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%lld", p + i);
  for (int i = 1; i <= n; ++i) scanf("%lld", ai + i), Add(i, ai[i]);
  Build(1, 0, n - 1);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a < 0) {
      a = -a;
      long long da = b - ai[a];
      ai[a] = b;
      Add(a, da);
      if (a <= n - 1) Upd(1, 0, n - 1, a, n - 1, (da + P) % P);
    } else {
      long long sl = Sum(a - 1), sr = Sum(b);
      int l = a, r = b - 1, pt = a - 1;
      while (l <= r) {
        long long sm = Sum(((l + r) >> 1));
        if (sm - sl < sr - sm)
          pt = ((l + r) >> 1), l = ((l + r) >> 1) + 1;
        else
          r = ((l + r) >> 1) - 1;
      }
      long long ans = 0;
      if (pt >= a) {
        ans = (ans + Query(1, 0, n - 1, a, pt)) % P;
        ans = (ans - sl % P * (p[pt + 1] - p[a] - (pt - a + 1)) % P + P) % P;
      }
      if (pt + 1 <= b - 1) {
        ans = (ans - Query(1, 0, n - 1, pt + 1, b - 1) + P) % P;
        ans = (ans + sr % P * (p[b] - p[pt + 1] - (b - 1 - pt))) % P;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}