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;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
using num = modnum<int(1e9) + 7>;
using ll = long long;
const int S = (1 << 18);
struct seg_node {
  ll sum_w;
  num sum_aw;
} seg[2 * S];
const int MAXN = 2.1e5;
int N;
int A[MAXN];
void recalc(int i) {
  assert(1 <= i && i < S);
  int l = 2 * i, r = 2 * i + 1;
  seg[i].sum_w = seg[l].sum_w + seg[r].sum_w;
  seg[i].sum_aw = seg[l].sum_aw + seg[r].sum_aw;
}
void update(int i, int nw) {
  assert(0 <= i && i <= N - 1);
  seg[S + i].sum_w = nw;
  seg[S + i].sum_aw = num(ll(A[i]) * nw);
  for (int a = (S + i) / 2; a; a /= 2) {
    recalc(a);
  }
}
seg_node get_sum(int l, int r) {
  seg_node res{0, 0};
  for (l += S, r += S; l < r; l /= 2, r /= 2) {
    if (l & 1) {
      res.sum_w += seg[l].sum_w;
      res.sum_aw += seg[l].sum_aw;
      l++;
    }
    if (r & 1) {
      r--;
      res.sum_w += seg[r].sum_w;
      res.sum_aw += seg[r].sum_aw;
    }
  }
  return res;
}
num solve(int l, int r) {
  assert(0 <= l && l <= r && r <= N - 1);
  r++;
  assert(0 <= l && l < r && r <= N);
  ll pl = get_sum(0, l).sum_w;
  ll pr = get_sum(0, r).sum_w;
  assert(pl < pr);
  ll pm = (pl + pr) / 2;
  assert(pl <= pm && pm < pr);
  int m = 1;
  while (m < S) {
    assert(seg[m].sum_w > pm);
    int ml = 2 * m;
    if (seg[ml].sum_w > pm) {
      m = ml;
    } else {
      pm -= seg[ml].sum_w;
      m = ml + 1;
    }
  }
  m -= S;
  assert(l <= m && m < r);
  seg_node sl = get_sum(l, m);
  seg_node sr = get_sum(m, r);
  return (num(A[m]) * num(sl.sum_w) - sl.sum_aw) +
         (sr.sum_aw - num(A[m]) * num(sr.sum_w));
}
int Q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> Q;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A[i] -= i;
  }
  for (int i = 0; i < N; i++) {
    int w;
    cin >> w;
    seg[S + i].sum_w = w;
    seg[S + i].sum_aw = num(ll(A[i]) * w);
  }
  for (int i = S - 1; i > 0; i--) {
    recalc(i);
  }
  for (int q = 0; q < Q; q++) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      int i = abs(x);
      i--;
      int nw = y;
      update(i, nw);
    } else {
      x--, y--;
      cout << solve(x, y) << '\n';
    }
  }
  return 0;
}