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;
inline long long read() {
  long long ans = 0, w = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') w = -1, c = getchar();
  while (isdigit(c)) ans = (ans << 1) + (ans << 3) + c - '0', c = getchar();
  return ans * w;
}
long long sum1[300010 << 2], sum2[300010 << 2];
void pu(int o) {
  sum1[o] = sum1[(o << 1)] + sum1[((o << 1) | 1)];
  sum2[o] = (sum2[(o << 1)] + sum2[((o << 1) | 1)]) % 1000000007;
}
int a[300010], w[300010];
int n, q;
long long mod(long long a) {
  a %= 1000000007;
  if (a < 0) a += 1000000007;
  return a;
}
long long add(long long a, long long b) { return mod(a + b); }
long long mul(long long a, long long b) { return mod(a * b); }
void pushup1(int t) { sum1[t] = sum1[(t << 1)] + sum1[(t << 1 | 1)]; }
void pushup2(int t) { sum2[t] = add(sum2[(t << 1)], sum2[(t << 1 | 1)]); }
void modify1(int t, int l, int r, int pos, long long vl) {
  if (l == r) {
    sum1[t] = vl;
    return;
  }
  if (pos <= ((l + r) >> 1))
    modify1((t << 1), l, ((l + r) >> 1), pos, vl);
  else
    modify1((t << 1 | 1), ((l + r) >> 1) + 1, r, pos, vl);
  pushup1(t);
}
void modify2(int t, int l, int r, int pos, long long vl) {
  if (l == r) {
    sum2[t] = vl;
    return;
  }
  if (pos <= ((l + r) >> 1))
    modify2((t << 1), l, ((l + r) >> 1), pos, vl);
  else
    modify2((t << 1 | 1), ((l + r) >> 1) + 1, r, pos, vl);
  pushup2(t);
}
long long qry1(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum1[o];
  long long ans = 0;
  if (((l + r) >> 1) >= ql) ans += qry1((o << 1), l, ((l + r) >> 1), ql, qr);
  if (((l + r) >> 1) < qr)
    ans += qry1(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
  return ans;
}
long long qry2(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum2[o];
  long long ans = 0;
  if (((l + r) >> 1) >= ql) ans = qry2((o << 1), l, ((l + r) >> 1), ql, qr);
  if (((l + r) >> 1) < qr)
    (ans += qry2(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr)) %= 1000000007;
  return ans;
}
int query1_pos(int t, int l, int r, int L, int R, long long vl) {
  if (l == r) return l;
  if (R <= ((l + r) >> 1))
    return query1_pos((t << 1), l, ((l + r) >> 1), L, R, vl);
  if (L > ((l + r) >> 1))
    return query1_pos((t << 1 | 1), ((l + r) >> 1) + 1, r, L, R, vl);
  long long suml = qry1((t << 1), l, ((l + r) >> 1), L, ((l + r) >> 1));
  if (suml >= vl)
    return query1_pos((t << 1), l, ((l + r) >> 1), L, ((l + r) >> 1), vl);
  else
    return query1_pos((t << 1 | 1), ((l + r) >> 1) + 1, r, ((l + r) >> 1) + 1,
                      R, vl - suml);
}
void build(int o, int l, int r) {
  if (l == r) {
    sum1[o] = w[l], sum2[o] = (long long)a[l] * w[l] % 1000000007;
    return;
  }
  build((o << 1), l, ((l + r) >> 1)),
      build(((o << 1) | 1), ((l + r) >> 1) + 1, r);
  pu(o);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), a[i] -= i;
  for (int i = 1; i <= n; ++i) scanf("%d", w + i);
  build(1, 1, n);
  while (q--) {
    int first = read(), second = read();
    if (first < 0) {
      first = -first;
      w[first] = second;
      modify1(1, 1, n, first, w[first]);
      modify2(1, 1, n, first, mul(a[first], w[first]));
    } else {
      long long sum = qry1(1, 1, n, first, second);
      int l = first, r = second, pos = -1;
      while (l <= r) {
        if (2 * qry1(1, 1, n, first, ((l + r) >> 1)) >= sum)
          pos = ((l + r) >> 1), r = ((l + r) >> 1) - 1;
        else
          l = ((l + r) >> 1) + 1;
      }
      long long ans = 0;
      ans = add(ans, mul(a[pos], add(qry1(1, 1, n, first, pos),
                                     -qry1(1, 1, n, pos, second))));
      ans =
          add(ans, add(-qry2(1, 1, n, first, pos), qry2(1, 1, n, pos, second)));
      printf("%lld\n", ans);
    }
  }
  return 0;
}