DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
long long A[maxn];
int n, m;
struct segTree {
  struct segNode {
    long long col;
    long long sum;
  } tree[maxn * 4];
  void push_up(int x) {
    tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
    if (tree[x << 1].col == tree[x << 1 | 1].col)
      tree[x].col = tree[x << 1].col;
    else
      tree[x].col = 0;
  }
  void push_down(int x, int l, int r) {
    if (tree[x].col) {
      long long change = tree[x].sum - tree[x << 1].sum - tree[x << 1 | 1].sum;
      change /= (r - l + 1);
      int mid = (l + r) >> 1;
      tree[x << 1].sum += change * (mid - l + 1);
      tree[x << 1 | 1].sum += change * (r - mid);
      tree[x << 1].col = tree[x << 1 | 1].col = tree[x].col;
    }
    return;
  }
  void build(int x, int l, int r) {
    if (l == r) {
      tree[x].sum = 0;
      tree[x].col = A[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(x << 1, l, mid);
    build(x << 1 | 1, mid + 1, r);
    push_up(x);
    return;
  }
  void update(int x, int l, int r, int ql, int qr, long long v) {
    if (l == r) {
      tree[x].sum += abs(tree[x].col - v);
      tree[x].col = v;
      return;
    }
    if (l == ql && r == qr && tree[x].col) {
      tree[x].sum += abs(tree[x].col - v) * 1LL * (r - l + 1);
      tree[x].col = v;
      return;
    }
    push_down(x, l, r);
    int mid = (l + r) >> 1;
    if (qr <= mid)
      update(x << 1, l, mid, ql, qr, v);
    else if (ql > mid)
      update(x << 1 | 1, mid + 1, r, ql, qr, v);
    else {
      update(x << 1, l, mid, ql, mid, v);
      update(x << 1 | 1, mid + 1, r, mid + 1, qr, v);
    }
    push_up(x);
    return;
  }
  long long query(int x, int l, int r, int ql, int qr) {
    if (l == ql && r == qr) return tree[x].sum;
    push_down(x, l, r);
    int mid = (l + r) >> 1;
    if (qr <= mid)
      return query(x << 1, l, mid, ql, qr);
    else if (ql > mid)
      return query(x << 1 | 1, mid + 1, r, ql, qr);
    else
      return query(x << 1, l, mid, ql, mid) +
             query(x << 1 | 1, mid + 1, r, mid + 1, qr);
  }
} T;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) A[i] = i;
  T.build(1, 1, n);
  int opr, a, b;
  long long c;
  while (m--) {
    scanf("%d %d %d", &opr, &a, &b);
    if (opr == 1) {
      scanf("%lld", &c);
      T.update(1, 1, n, a, b, c);
    } else
      printf("%lld\n", T.query(1, 1, n, a, b));
  }
  return 0;
}
