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 = 100010;
const int INF = 0x7f7f7f7f;
const double pi = 4.0 * atan(1.0);
int cases;
struct Segment_Tree {
  long long lazy[MAXN << 2];
  long long data[MAXN << 2];
  long long flag[MAXN << 2];
  void pushup(int o) {
    data[o] = data[(o << 1)] + data[(o << 1) | 1];
    if (flag[(o << 1)] != flag[(o << 1) | 1]) {
      flag[o] = 0;
    } else {
      flag[o] = flag[(o << 1)];
    }
  }
  void setlazy(int o, int m, long long v) {
    lazy[o] += v;
    data[o] += m * v;
  }
  void pushdown(int o, int m) {
    if (lazy[o]) {
      setlazy((o << 1), m - (m >> 1), lazy[o]);
      setlazy((o << 1) | 1, m >> 1, lazy[o]);
      flag[(o << 1)] = flag[(o << 1) | 1] = flag[o];
      lazy[o] = 0;
    }
  }
  void build_tree(int o, int l, int r) {
    if (l == r) {
      flag[o] = l;
      return;
    }
    int mid = (l + r) >> 1;
    build_tree((o << 1), l, mid);
    build_tree((o << 1) | 1, mid + 1, r);
    pushup(o);
  }
  void update(int o, int l, int r, int x, int y, long long v) {
    if (x <= l && r <= y && flag[o] != 0) {
      lazy[o] += abs(flag[o] - v);
      data[o] += (long long)(r - l + 1) * abs(flag[o] - v);
      flag[o] = v;
      return;
    }
    pushdown(o, r - l + 1);
    int mid = (l + r) >> 1;
    if (x <= mid) update((o << 1), l, mid, x, y, v);
    if (y > mid) update((o << 1) | 1, mid + 1, r, x, y, v);
    pushup(o);
  }
  long long query(int o, int l, int r, int x, int y) {
    if (x <= l && r <= y) {
      return data[o];
    }
    pushdown(o, r - l + 1);
    int mid = (l + r) >> 1;
    if (y <= mid) return query((o << 1), l, mid, x, y);
    if (x > mid) return query((o << 1) | 1, mid + 1, r, x, y);
    return query((o << 1), l, mid, x, y) +
           query((o << 1) | 1, mid + 1, r, x, y);
  }
} segtree;
int main() {
  int n, m, op, x, y;
  long long z;
  scanf("%d%d", &n, &m);
  segtree.build_tree(1, 1, n);
  for (int i = 0; i < m; i++) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%lld", &x, &y, &z);
      segtree.update(1, 1, n, x, y, z);
    } else if (op == 2) {
      scanf("%d%d", &x, &y);
      printf("%lld\n", segtree.query(1, 1, n, x, y));
    }
  }
}
