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 gcd(long long a, long long b) {
  if (a == 0LL) return b;
  if (b == 0LL) return a;
  return gcd(b, a % b);
}
const int N = 200005;
long long A[N], W[N];
pair<long long, long long> tree[N << 2];
pair<long long, long long> temp[N << 2];
void build(int node, int l, int r) {
  if (l == r) {
    tree[node] = make_pair(W[l], (A[l] * W[l]) % 1000000007LL);
    return;
  }
  int mid = (l + r) >> 1;
  int left = node << 1;
  int right = left + 1;
  build(left, l, mid);
  build(right, mid + 1, r);
  tree[node].first = tree[left].first + tree[right].first;
  tree[node].second = (tree[left].second + tree[right].second) % 1000000007LL;
  return;
}
void update(int node, int l, int r, int idx) {
  if (l == r) {
    tree[node] = make_pair(W[idx], (A[idx] * W[idx]) % 1000000007LL);
    return;
  }
  int mid = (l + r) >> 1;
  int left = node << 1;
  int right = left + 1;
  if (idx <= mid)
    update(left, l, mid, idx);
  else
    update(right, mid + 1, r, idx);
  tree[node].first = tree[left].first + tree[right].first;
  tree[node].second = (tree[left].second + tree[right].second) % 1000000007LL;
  return;
}
pair<long long, long long> query(int node, int l, int r, int a, int b) {
  if (l == a && r == b) {
    temp[node] = tree[node];
    return tree[node];
  }
  int mid = (l + r) >> 1;
  int left = node << 1;
  int right = left + 1;
  if (b <= mid) {
    temp[node] = query(left, l, mid, a, b);
    return temp[node];
  } else if (a > mid) {
    temp[node] = query(right, mid + 1, r, a, b);
    return temp[node];
  }
  pair<long long, long long> L = query(left, l, mid, a, mid);
  pair<long long, long long> R = query(right, mid + 1, r, mid + 1, b);
  pair<long long, long long> ans =
      make_pair(L.first + R.first, (L.second + R.second) % 1000000007LL);
  temp[node] = ans;
  return ans;
}
int query1(int node, int l, int r, int a, int b, long long val) {
  int mid = (l + r) >> 1;
  int left = node << 1;
  int right = left + 1;
  if (l == a && r == b) {
    if (l == r) return l;
    if (2LL * tree[left].first < val)
      return query1(right, mid + 1, r, mid + 1, r,
                    val - 2LL * tree[left].first);
    return query1(left, l, mid, l, mid, val);
  }
  if (b <= mid)
    return query1(left, l, mid, a, b, val);
  else if (a > mid)
    return query1(right, mid + 1, r, a, b, val);
  long long val1 = temp[left].first;
  if (2LL * val1 < val)
    return query1(right, mid + 1, r, mid + 1, b, val - 2LL * val1);
  return query1(left, l, mid, a, mid, val);
}
long long getAns(int n, int L, int R, int ind) {
  long long X = A[ind];
  long long ans = 0LL;
  pair<long long, long long> val = query(1, 1, n, L, ind);
  val.first = val.first % 1000000007LL;
  ans = (ans + ((X * val.first) % 1000000007LL) - val.second +
         2LL * 1000000007LL) %
        1000000007LL;
  val = query(1, 1, n, ind, R);
  val.first = val.first % 1000000007LL;
  ans = (ans + val.second - ((X * val.first) % 1000000007LL) +
         2LL * 1000000007LL) %
        1000000007LL;
  return ans;
}
void solve1(int n, int L, int R) {
  pair<long long, long long> val = query(1, 1, n, L, R);
  int ind = query1(1, 1, n, L, R, val.first);
  cout << getAns(n, L, R, ind) << "\n";
  return;
}
void solve() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    A[i] = A[i] - i;
  }
  for (int i = 1; i <= n; i++) cin >> W[i];
  build(1, 1, n);
  int L, R;
  while (q--) {
    cin >> L >> R;
    if (L < 0) {
      L = -1 * L;
      W[L] = R;
      update(1, 1, n, L);
      continue;
    }
    solve1(n, L, R);
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  clock_t clk = clock();
  int t = 1;
  for (int tests = 1; tests <= t; tests++) {
    solve();
  }
  clk = clock() - clk;
  cerr << "Time Elapsed: " << fixed << setprecision(10)
       << ((long double)clk) / CLOCKS_PER_SEC << "\n";
  return 0;
}