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>
#pragma GCC optimize(3)
using namespace std;
long long n, q, a[200005], p[200005];
long long _mod(long long x) { return x >= 1000000007 ? x - 1000000007 : x; }
struct segtree {
  long long sum[4 * 200005];
  long long sum2[4 * 200005], sum3[4 * 200005];
  void pushup(long long k) {
    sum[k] = sum[k * 2] + sum[k * 2 + 1];
    sum2[k] = _mod(sum2[k * 2] + sum2[k * 2 + 1]);
    sum3[k] = _mod(sum3[k * 2] + sum3[k * 2 + 1]);
  }
  void build(long long k, long long l, long long r) {
    if (l == r) {
      sum[k] = a[l];
      sum2[k] = 1LL * a[l] * p[l] % 1000000007;
      sum3[k] = 1LL * l * a[l] % 1000000007;
      return;
    }
    long long mid = (l + r) / 2;
    build(k * 2, l, mid);
    build(k * 2 + 1, mid + 1, r);
    pushup(k);
  }
  void update(long long k, long long l, long long r, long long pos,
              long long v) {
    if (l == r) {
      sum[k] = v;
      sum2[k] = 1LL * v * p[l] % 1000000007;
      sum3[k] = 1LL * v * l % 1000000007;
      return;
    }
    long long mid = (l + r) / 2;
    if (pos <= mid)
      update(k * 2, l, mid, pos, v);
    else
      update(k * 2 + 1, mid + 1, r, pos, v);
    pushup(k);
  }
  long long query1(long long k, long long l, long long r, long long x,
                   long long y) {
    if (x > r || l > y) return 0;
    if (l >= x && r <= y) return sum[k];
    long long mid = (l + r) / 2;
    return query1(k * 2, l, mid, x, y) + query1(k * 2 + 1, mid + 1, r, x, y);
  }
  long long find(long long k, long long l, long long r, long long s) {
    if (l == r) return l;
    long long mid = (l + r) / 2;
    if (sum[k * 2] <= s) return find(k * 2 + 1, mid + 1, r, s - sum[k * 2]);
    return find(k * 2, l, mid, s);
  }
  long long query2(long long k, long long l, long long r, long long x,
                   long long y) {
    if (x > r || l > y) return 0;
    if (l >= x && r <= y) return sum2[k];
    long long mid = (l + r) / 2;
    return _mod(query2(k * 2, l, mid, x, y) +
                query2(k * 2 + 1, mid + 1, r, x, y));
  }
  long long query3(long long k, long long l, long long r, long long x,
                   long long y) {
    if (x > r || l > y) return 0;
    if (l >= x && r <= y) return sum3[k];
    long long mid = (l + r) / 2;
    return _mod(query3(k * 2, l, mid, x, y) +
                query3(k * 2 + 1, mid + 1, r, x, y));
  }
} seg;
int main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &p[i]);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  seg.build(1, 1, n);
  long long l, r;
  for (long long i = 0; i < q; i++) {
    scanf("%lld%lld", &l, &r);
    if (l < 0)
      seg.update(1, 1, n, -l, r);
    else {
      long long A = seg.query1(1, 1, n, 1, l - 1),
                B = seg.query1(1, 1, n, l, r), s = A + B / 2;
      long long pos = seg.find(1, 1, n, s);
      long long ans = 0;
      ans = (1LL * (p[pos] - pos) *
                 (seg.query1(1, 1, n, l, pos - 1) % 1000000007) +
             seg.query3(1, 1, n, l, pos - 1) - seg.query2(1, 1, n, l, pos - 1) +
             1000000007) %
            1000000007;
      ans = (ans + seg.query2(1, 1, n, pos + 1, r) +
             1LL * (pos - p[pos]) *
                 (seg.query1(1, 1, n, pos + 1, r) % 1000000007) % 1000000007 -
             seg.query3(1, 1, n, pos + 1, r) + 1000000007) %
            1000000007;
      if (ans < 0) ans += 1000000007;
      printf("%lld\n", ans);
    }
  }
  return 0;
}