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 Val(long long val) {
  return (val % 1000000007 + 1000000007) % 1000000007;
}
long long n, q;
long long pos[264000], cost[264000];
struct node {
  long long lpos, rpos, num;
  long long lsum, rsum;
  long long cost;
} seg[264000 << 1];
inline node merge(node lch, node rch) {
  node ret;
  ret.lpos = lch.lpos;
  ret.rpos = rch.rpos;
  ret.num = lch.num + rch.num;
  ret.lsum =
      Val(lch.lsum + rch.lsum + (rch.lpos - (lch.lpos + lch.num)) * rch.cost);
  ret.rsum =
      Val(rch.rsum + lch.rsum + ((rch.rpos - rch.num) - lch.rpos) * lch.cost);
  ret.cost = Val(lch.cost + rch.cost);
  return ret;
}
void build(long long l, long long r, long long p) {
  if (l == r)
    seg[p] = node{pos[l], pos[l], 1, 0, 0, cost[l]};
  else {
    long long mid = (l + r) >> 1, lch = p << 1, rch = p << 1 | 1;
    build(l, mid, lch);
    build(mid + 1, r, rch);
    seg[p] = merge(seg[lch], seg[rch]);
  }
}
void modify(long long l, long long r, long long idx, long long p) {
  if (l == r)
    seg[p] = node{pos[l], pos[l], 1, 0, 0, cost[l]};
  else {
    long long mid = (l + r) >> 1, lch = p << 1, rch = p << 1 | 1;
    if (idx <= mid)
      modify(l, mid, idx, lch);
    else
      modify(mid + 1, r, idx, rch);
    seg[p] = merge(seg[lch], seg[rch]);
  }
}
node query(long long l, long long r, long long ql, long long qr, long long p) {
  if (l == ql && r == qr)
    return seg[p];
  else {
    long long mid = (l + r) >> 1, lch = p << 1, rch = p << 1 | 1;
    if (qr <= mid)
      return query(l, mid, ql, qr, lch);
    else if (ql > mid)
      return query(mid + 1, r, ql, qr, rch);
    else
      return merge(query(l, mid, ql, mid, lch),
                   query(mid + 1, r, mid + 1, qr, rch));
  }
}
long long bit[264000];
inline long long lowbit(long long x) { return x & -x; }
void add(long long idx, long long val) {
  while (idx <= n) {
    bit[idx] += val;
    idx += lowbit(idx);
  }
}
long long sum(long long idx) {
  long long ret = 0;
  while (idx > 0) {
    ret += bit[idx];
    idx -= lowbit(idx);
  }
  return ret;
}
long long query(long long l, long long r) {
  return sum(r) - (l == 1 ? 0 : sum(l - 1));
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> pos[i];
  for (long long i = 1; i <= n; i++) cin >> cost[i];
  build(1, n, 1);
  for (long long i = 1; i <= n; i++) add(i, cost[i]);
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      add(-x, -cost[-x]);
      cost[-x] = y;
      add(-x, cost[-x]);
      modify(1, n, -x, 1);
    } else {
      if (x == y) {
        cout << 0 << '\n';
      } else {
        long long l = x, r = y, m;
        long long tot = query(x, y);
        while (true) {
          if (r - l < 3) {
            for (m = r; m >= l && 2ll * query(x, m) >= tot; m--)
              ;
            m++;
            break;
          }
          m = (l + r) >> 1;
          if (2ll * query(x, m) >= tot)
            r = m;
          else
            l = m + 1;
        }
        cout << Val(query(1, n, x, m, 1).rsum + query(1, n, m, y, 1).lsum)
             << '\n';
      }
    }
  }
  return 0;
}