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 long long MOD = 1000000007;
const long long MAX = 200007;
long long sum(long long a, long long b) {
  return a + b >= MOD ? a + b - MOD : a + b;
}
long long sub(long long a, long long b) {
  return a - b < 0 ? a - b + MOD : a - b;
}
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long ps[MAX];
long long ar[MAX];
long long tr[4 * MAX];
long long tr2[4 * MAX];
long long tr3[4 * MAX];
void init(long long u, long long l, long long r) {
  if (l == r) {
    tr[u] = ar[l];
    tr2[u] = mul(ar[l], ps[l]);
    tr3[u] = mul(ar[l], l);
    return;
  }
  long long mid = (l + r) / 2;
  init(u * 2, l, mid);
  init(u * 2 + 1, mid + 1, r);
  tr[u] = tr[u * 2] + tr[u * 2 + 1];
  tr2[u] = sum(tr2[u * 2], tr2[u * 2 + 1]);
  tr3[u] = sum(tr3[u * 2], tr3[u * 2 + 1]);
}
void update(long long u, long long l, long long r, long long idx, long long v) {
  if (l == r) {
    ar[l] = v;
    tr[u] = ar[l];
    tr2[u] = mul(ar[l], ps[l]);
    tr3[u] = mul(ar[l], l);
    return;
  }
  long long mid = (l + r) / 2;
  if (idx <= mid)
    update(u * 2, l, mid, idx, v);
  else
    update(u * 2 + 1, mid + 1, r, idx, v);
  tr[u] = tr[u * 2] + tr[u * 2 + 1];
  tr2[u] = sum(tr2[u * 2], tr2[u * 2 + 1]);
  tr3[u] = sum(tr3[u * 2], tr3[u * 2 + 1]);
}
pair<long long, long long> binSearch(long long u, long long l, long long r,
                                     long long x, long long y, long long sm,
                                     long long lft = 0) {
  if (r < x || y < l) return pair<long long, long long>(0, 0);
  if (x <= l && r <= y) {
    if (lft + tr[u] <= sm / 2 || l == r)
      return pair<long long, long long>(r, lft + tr[u]);
  }
  long long mid = (l + r) / 2;
  pair<long long, long long> _l = binSearch(u * 2, l, mid, x, y, sm, lft);
  if (_l.second <= sm / 2)
    return binSearch(u * 2 + 1, mid + 1, r, x, y, sm, _l.second);
  else
    return _l;
}
long long query(long long u, long long l, long long r, long long x,
                long long y) {
  if (x > y) return 0;
  if (r < x || y < l) return 0;
  if (x <= l && r <= y) return tr[u];
  long long mid = (l + r) / 2;
  return query(u * 2, l, mid, x, y) + query(u * 2 + 1, mid + 1, r, x, y);
}
long long query2(long long u, long long l, long long r, long long x,
                 long long y) {
  if (x > y) return 0;
  if (r < x || y < l) return 0;
  if (x <= l && r <= y) return tr2[u];
  long long mid = (l + r) / 2;
  return sum(query2(u * 2, l, mid, x, y), query2(u * 2 + 1, mid + 1, r, x, y));
}
long long query3(long long u, long long l, long long r, long long x,
                 long long y) {
  if (x > y) return 0;
  if (r < x || y < l) return 0;
  if (x <= l && r <= y) return tr3[u];
  long long mid = (l + r) / 2;
  return sum(query3(u * 2, l, mid, x, y), query3(u * 2 + 1, mid + 1, r, x, y));
}
long long n;
long long wut(long long l, long long r, long long x) {
  long long ans = 0;
  long long leftsum = query(1, 1, n, l, x) % MOD;
  long long rightsum = query(1, 1, n, x, r) % MOD;
  long long leftsum2 = query2(1, 1, n, l, x);
  long long rightsum2 = query2(1, 1, n, x, r);
  long long leftsum3 = query3(1, 1, n, l, x);
  long long rightsum3 = query3(1, 1, n, x, r);
  ans = sum(ans, sub(mul(ps[x], leftsum), leftsum2));
  ans = sub(ans, sub(mul(x, leftsum), leftsum3));
  ans = sum(ans, sub(rightsum2, mul(ps[x], rightsum)));
  ans = sub(ans, sub(rightsum3, mul(x, rightsum)));
  return ans;
}
long long solve(long long l, long long r) {
  if (l == r) return 0;
  long long sm = query(1, 1, n, l, r);
  long long y = binSearch(1, 1, n, l, r, sm).first;
  long long x = y - 1;
  long long leftsum = query(1, 1, n, l, x);
  long long rightsum = query(1, 1, n, y, r);
  if (leftsum <= rightsum) return wut(l, r, y);
  return wut(l, r, x);
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> ps[i];
  for (long long i = 1; i <= n; i++) cin >> ar[i];
  init(1, 1, n);
  while (q--) {
    long long a, b;
    cin >> a >> b;
    if (a < 0)
      update(1, 1, n, -a, b);
    else
      cout << solve(a, b) << '\n';
  }
  return 0;
}