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]. 
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long double PI = acosl(-1);
const long long infl = 1e18 + 0;
const int inf = 1e9;
const int nmax = 2e5 + 5;
const int MAXLG = log2(nmax) + 1;
long long koi[nmax];
long long w[nmax];
long long n;
long long bst[4 * nmax];
long long wkoi[4 * nmax];
void bild(long long a, long long L, long long R) {
  if (L == R) {
    bst[a] = w[L];
    wkoi[a] = koi[L] * w[L] % mod;
    return;
  }
  long long mid = (L + R) / 2;
  bild(2 * a, L, mid);
  bild(2 * a + 1, mid + 1, R);
  bst[a] = bst[a * 2] + bst[a * 2 + 1];
  wkoi[a] = (wkoi[a * 2] + wkoi[a * 2 + 1]) % mod;
}
long long getpos(long long a, long long L, long long R, long long dorkar) {
  if (L == R) return L;
  long long mid = (L + R) / 2;
  if (bst[2 * a] >= dorkar) return getpos(2 * a, L, mid, dorkar);
  return getpos(2 * a + 1, mid + 1, R, dorkar - bst[2 * a]);
}
void update(long long a, long long L, long long R, long long pos) {
  if (L == R) {
    bst[a] = w[L];
    wkoi[a] = koi[L] * w[L] % mod;
    return;
  }
  long long mid = (L + R) / 2;
  if (pos <= mid)
    update(2 * a, L, mid, pos);
  else
    update(2 * a + 1, mid + 1, R, pos);
  bst[a] = bst[a * 2] + bst[a * 2 + 1];
  wkoi[a] = (wkoi[a * 2] + wkoi[a * 2 + 1]) % mod;
}
long long getsum(long long a, long long L, long long R, long long l,
                 long long r) {
  if (l > R or L > r or l > r) return 0;
  if (l <= L and R <= r) return bst[a];
  long long mid = (L + R) / 2;
  return getsum(2 * a, L, mid, l, r) + getsum(2 * a + 1, mid + 1, R, l, r);
}
long long getwkoi(long long a, long long L, long long R, long long l,
                  long long r) {
  if (l > R or L > r or l > r) return 0;
  if (l <= L and R <= r) return wkoi[a];
  long long mid = (L + R) / 2;
  return (getwkoi(2 * a, L, mid, l, r) + getwkoi(2 * a + 1, mid + 1, R, l, r)) %
         mod;
}
void updeet(long long pos, long long niuw) {
  w[pos] = niuw;
  update(1, 1, n, pos);
}
long long bamcost(long long L, long long R) {
  long long renj = R - L + 1;
  long long W = getsum(1, 1, n, L, R) % mod;
  long long W_koi = getwkoi(1, 1, n, L, R);
  long long costo = (W * koi[R] % mod - W_koi + mod) % mod;
  return costo;
}
long long dancost(long long L, long long R) {
  long long renj = R - L + 1;
  long long W = getsum(1, 1, n, L, R) % mod;
  long long W_koi = getwkoi(1, 1, n, L, R);
  long long costo = (-W * koi[L] % mod + W_koi + mod) % mod;
  return costo;
}
long long kuery(long long L, long long R) {
  long long sam = getsum(1, 1, n, L, R);
  long long pos = getpos(1, 1, n, (sam + 1) / 2 + getsum(1, 1, n, 1, L - 1));
  long long b = bamcost(L, pos), d = dancost(pos, R);
  return (b + d) % mod;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> koi[i], koi[i] -= i;
  for (long long i = 1; i <= n; i++) cin >> w[i];
  bild(1, 1, n);
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      updeet(-x, y);
    } else {
      cout << kuery(x, y) << "\n";
    }
  }
}
