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 int MAXN = 200000 + 100;
const int MOD = 1e9 + 7;
const long long MODM = static_cast<long long>(MOD) * 100;
int w[MAXN], x[MAXN];
int N, Q;
int prod(int a, int b) {
  if (a > 0 xor b > 0)
    return -static_cast<int>((static_cast<long long>(-a) * b) % MOD);
  return static_cast<int>((static_cast<long long>(a) * b) % MOD);
}
int sum(int a, int b) {
  return static_cast<int>((static_cast<long long>(a) + b + MODM) % MOD);
}
int dif(int a, int b) {
  return static_cast<int>((static_cast<long long>(a) - b + MODM) % MOD);
}
struct NODE {
 public:
  int wl, wr;
  int ws;
  NODE* l;
  NODE* r;
  NODE() {
    l = nullptr;
    r = nullptr;
  }
};
struct ST {
 public:
  NODE* h;
  int s;
  ST() { h = nullptr; }
  ST(int v) {
    s = v;
    h = new NODE();
    init(h, 0, s);
  }
  void init(NODE* n, int nl, int nr) {
    n->wl = n->wr = n->ws = 0;
    for (int i = nl; i < nr; i++) {
      n->ws = sum(n->ws, w[i]);
      n->wl = sum(n->wl, prod(w[i], x[i] - x[nl] - (i - nl)));
      n->wr = sum(n->wr, prod(w[i], x[nr - 1] - x[i] - (nr - 1 - i)));
    }
    if (nr - nl > 1) {
      int m = nl + (nr - nl) / 2;
      n->l = new NODE();
      init(n->l, nl, m);
      n->r = new NODE();
      init(n->r, m, nr);
    }
  }
  void upd(int id, int nw) { upd(h, 0, s, id, nw); }
  void upd(NODE* n, int nl, int nr, int id, int nw) {
    assert(nl <= id and id < nr);
    n->ws = sum(n->ws, dif(nw, w[id]));
    n->wl = sum(n->wl, prod(x[id] - x[nl] - (id - nl), dif(nw, w[id])));
    n->wr = sum(n->wr, prod(x[nr - 1] - x[id] - (nr - 1 - id), dif(nw, w[id])));
    if (nr - nl > 1) {
      int m = nl + (nr - nl) / 2;
      if (id < m) upd(n->l, nl, m, id, nw);
      if (id >= m) upd(n->r, m, nr, id, nw);
    }
  }
  int ql(int gl, int gr, int g) { return ql(h, 0, s, gl, gr, g); }
  int ql(NODE* n, int nl, int nr, int gl, int gr, int g) {
    if (gl >= gr or gr <= nl or nr <= gl) return 0;
    assert(g < nr);
    if (g <= nl and gl <= nl and gr >= nr)
      return sum(n->wl, prod(n->ws, x[nl] - x[g] - (nl - g)));
    int ret = 0;
    if (nr - nl > 1) {
      int m = nl + (nr - nl) / 2;
      if (nr > m) ret = ql(n->r, m, nr, gl, gr, g);
      if (nl < m and g < m) ret = sum(ret, ql(n->l, nl, m, gl, gr, g));
    }
    return ret;
  }
  int qr(int gl, int gr, int g) { return qr(h, 0, s, gl, gr, g); }
  int qr(NODE* n, int nl, int nr, int gl, int gr, int g) {
    if (gl >= gr or nl >= gr or gl >= nr) return 0;
    assert(g >= nl);
    if (g >= nr and gl <= nl and gr >= nr)
      return sum(n->wr, prod(n->ws, x[g] - x[nr - 1] - (g - nr + 1)));
    int ret = 0;
    if (nr - nl > 1) {
      int m = nl + (nr - nl) / 2;
      if (nl < m) ret = qr(n->l, nl, m, gl, gr, g);
      if (nr > m and g >= m) ret = sum(ret, qr(n->r, m, nr, gl, gr, g));
    }
    return ret;
  }
} st;
struct BIT {
  long long arr[MAXN];
  int s, l2s;
  BIT() { s = 0; }
  BIT(int v) {
    s = v;
    for (l2s = 1; l2s <= s; l2s <<= 1)
      ;
    for (int i = 1; i <= s; i++) arr[i] = 0;
    for (int i = 0; i < s; i++) upd(i, 2 * w[i]);
  }
  void upd(int id, int nw) {
    for (int i = id + 1; i <= s; i += (i & (-i))) arr[i] += nw - w[id];
  }
  long long qry(int id) {
    long long ret = 0;
    for (int i = id + 1; i > 0; i -= (i & (-i))) ret += arr[i];
    return ret;
  }
  int srh(long long goal) {
    int id = 0;
    long long cur = 0;
    for (int i = l2s; i > 0; i >>= 1) {
      if (id + i > s) continue;
      if (cur + arr[id + i] < goal) {
        id += i;
        cur += arr[id];
      }
    }
    return id;
  }
} psm;
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 0; i < N; i++) scanf("%d", x + i);
  for (int i = 0; i < N; i++) scanf("%d", w + i);
  st = ST(N);
  psm = BIT(N);
  int a, b;
  for (int i = 0; i < Q; i++) {
    scanf("%d%d", &a, &b);
    if (a < 0) {
      a = -a - 1;
      st.upd(a, b);
      psm.upd(a, b);
      w[a] = b;
    } else {
      a--;
      long long aw = psm.qry(a - 1);
      long long bw = psm.qry(b - 1);
      long long md = aw + (bw - aw + 1) / 2;
      int m = psm.srh(md);
      int ans = sum(st.ql(m, b, m), st.qr(a, m, m));
      printf("%d\n", ans);
    }
  }
  return 0;
}