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;
inline int scan() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x;
}
const int N = 100001;
const int SN = 400001;
long long segtree[SN] = {0};
long long color[SN] = {0};
long long lazy[SN] = {0};
int n, m;
void build(int l, int r, int node) {
  if (l == r) {
    color[node] = l;
  } else {
    int lc = node << 1;
    int rc = lc | 1;
    int mid = (l + r) >> 1;
    build(l, mid, lc);
    build(mid + 1, r, rc);
  }
}
void push(int l, int r, int node) {
  if (!lazy[node]) {
    return;
  }
  segtree[node] += 1LL * (r - l + 1) * lazy[node];
  if (l != r) {
    int lc = node << 1;
    int rc = lc | 1;
    lazy[lc] += lazy[node];
    lazy[rc] += lazy[node];
  }
  lazy[node] = 0;
}
void godown(int l, int r, int node, int val) {
  push(l, r, node);
  if (color[node] > 0) {
    lazy[node] += abs(color[node] - val);
    push(l, r, node);
    color[node] = val;
  } else {
    int lc = node << 1;
    int rc = lc | 1;
    int mid = (l + r) >> 1;
    godown(l, mid, lc, val);
    godown(mid + 1, r, rc, val);
    segtree[node] = segtree[lc] + segtree[rc];
    color[node] = val;
  }
}
void update(int l, int r, int node, int ql, int qr, int val) {
  int lc = node << 1;
  int rc = lc | 1;
  int mid = (l + r) >> 1;
  if (color[node] > 0 && l != r) {
    color[lc] = color[node];
    color[rc] = color[node];
  }
  push(l, r, node);
  if (l > qr || r < ql) {
    return;
  }
  if (l >= ql && r <= qr) {
    godown(l, r, node, val);
    return;
  }
  update(l, mid, lc, ql, qr, val);
  update(mid + 1, r, rc, ql, qr, val);
  if (color[lc] == color[rc]) {
    color[node] = color[lc];
  } else {
    color[node] = 0;
  }
  segtree[node] = segtree[lc] + segtree[rc];
}
long long query(int l, int r, int node, int ql, int qr) {
  push(l, r, node);
  if (l > qr || r < ql) {
    return 0;
  }
  if (l >= ql && r <= qr) {
    return segtree[node];
  }
  int lc = node << 1;
  int rc = lc | 1;
  int mid = (l + r) >> 1;
  return query(l, mid, lc, ql, qr) + query(mid + 1, r, rc, ql, qr);
}
int main() {
  n = scan(), m = scan();
  build(1, n, 1);
  while (m--) {
    int type = scan(), l = scan(), r = scan();
    if (type & 1) {
      int val = scan();
      update(1, n, 1, l, r, val);
    } else {
      printf("%lld\n", query(1, n, 1, l, r));
    }
  }
}
