AtCoDeer the deer found N rectangle lying on the table, each with height 1.
If we consider the surface of the desk as a two-dimensional plane, the i-th rectangle i(1≤i≤N) covers the vertical range of [i-1,i] and the horizontal range of [l_i,r_i], as shown in the following figure:

AtCoDeer will move these rectangles horizontally so that all the rectangles are connected.
For each rectangle, the cost to move it horizontally by a distance of x, is x.
Find the minimum cost to achieve connectivity.
It can be proved that this value is always an integer under the constraints of the problem.

-----Constraints-----
 - All input values are integers.
 - 1≤N≤10^5
 - 1≤l_i<r_i≤10^9

-----Partial Score-----
 - 300 points will be awarded for passing the test set satisfying 1≤N≤400 and 1≤l_i<r_i≤400.

-----Input-----
The input is given from Standard Input in the following format:
N
l_1 r_1
l_2 r_2
:
l_N r_N

-----Output-----
Print the minimum cost to achieve connectivity.

-----Sample Input-----
3
1 3
5 7
1 3

-----Sample Output-----
2

The second rectangle should be moved to the left by a distance of 2.

Difficulty: competition

Starter code:

Solutions:
# seishin.py
N = int(input())
P = [list(map(int, input().split())) for i in range(N)]

from heapq import heappush, heappop

l0, r0 = P[0]

L = [-l0+1]
R = [l0-1]
s = t = 0

res = 0
for i in range(N-1):
    l0, r0 = P[i]
    l1, r1 = P[i+1]
    s += (r1 - l1); t += (r0 - l0)
    if -s-L[0] <= l1-1 <= t+R[0]:
        heappush(L, -l1+1-s)
        heappush(R, l1-1-t)
    elif l1-1 < -s-L[0]:
        heappush(L, -l1+1-s)
        heappush(L, -l1+1-s)
        p = -heappop(L)-s
        heappush(R, p-t)
        res += (p - (l1-1))
    elif t+R[0] < l1-1:
        heappush(R, l1-1-t)
        heappush(R, l1-1-t)
        p = heappop(R) + t
        heappush(L, -p-s)
        res += ((l1-1) - p)
print(res)

N = int(input())
P = [list(map(int, input().split())) for i in range(N)]
 
INF = 10**18
 
from heapq import heappush, heappop
 
l0, r0 = P[0]
 
L = [-l0+1]
R = [l0-1]
s = t = 0
 
res = 0
for i in range(N-1):
    l0, r0 = P[i]
    l1, r1 = P[i+1]
    s += (r1 - l1); t += (r0 - l0)
    if -s-L[0] <= l1-1 <= t+R[0]:
        heappush(L, -l1+1-s)
        heappush(R, l1-1-t)
    elif l1-1 < -s-L[0]:
        heappush(L, -l1+1-s)
        heappush(L, -l1+1-s)
        p = -heappop(L)-s
        heappush(R, p-t)
        res += (p - (l1-1))
    elif t+R[0] < l1-1:
        heappush(R, l1-1-t)
        heappush(R, l1-1-t)
        p = heappop(R) + t
        heappush(L, -p-s)
        res += (l1-1 - p)
print(res)

# seishin.py
N = int(input())
P = [list(map(int, input().split())) for i in range(N)]

INF = 10**18

from heapq import heappush, heappop

l0, r0 = P[0]

L = [-l0+1]
R = [l0-1]
s = t = 0

res = 0
for i in range(N-1):
    l0, r0 = P[i]
    l1, r1 = P[i+1]
    s += (r1 - l1); t += (r0 - l0)
    if -s-L[0] <= l1-1 <= t+R[0]:
        heappush(L, -l1+1-s)
        heappush(R, l1-1-t)
    elif l1-1 < -s-L[0]:
        heappush(L, -l1+1-s)
        heappush(L, -l1+1-s)
        p = -heappop(L)-s
        heappush(R, p-t)
        res += (p - (l1-1))
    elif t+R[0] < l1-1:
        heappush(R, l1-1-t)
        heappush(R, l1-1-t)
        p = heappop(R) + t
        heappush(L, -p-s)
        res += (l1-1 - p)
print(res)

# seishin.py
N = int(input())
P = [list(map(int, input().split())) for i in range(N)]

INF = 10**18

from heapq import heappush, heappop

l0, r0 = P[0]

L = [-l0+1]
R = [l0-1]
s = t = 0

def debug(L, s, t, R):
    L0 = L[:]
    Q1 = []; Q2 = []
    while L0:
        Q1.append(-s-heappop(L0))
    R0 = R[:]
    while R0:
        Q2.append(t+heappop(R0))
    print(("debug:", *Q1[::-1]+Q2))


#print(L, s, t, R)
res = 0
for i in range(N-1):
    l0, r0 = P[i]
    l1, r1 = P[i+1]
    #print(">", l1, r1)
    s += (r1 - l1); t += (r0 - l0)
    if -s-L[0] <= l1-1 <= t+R[0]:
        #print(0)
        heappush(L, -l1+1-s)
        heappush(R, l1-1-t)
        # res += 0
    elif l1-1 < -s-L[0]:
        #print(1)
        heappush(L, -l1+1-s)
        heappush(L, -l1+1-s)
        p = -heappop(L)-s
        #d = (-L[0]-s) - p
        heappush(R, p-t)
        #print(d)
        #res += d
        res += (p - (l1-1))
    elif t+R[0] < l1-1:
        #print(2)
        heappush(R, l1-1-t)
        heappush(R, l1-1-t)
        p = heappop(R) + t
        #d = R[0]+t - p
        heappush(L, -p-s)
        #print(d)
        res += (l1-1 - p)
    #print(L, s, t, R, -s-L[0], R[0]+t, res)
    #debug(L, s, t, R)
print(res)


import sys
input = sys.stdin.readline
from heapq import heappop, heappush

"""
f(x) = （一番上の長方形の左端がxに来るときのコストの最小値） を関数ごと更新していきたい
更新後をg(x)とする
g(x) = |x-L| + min_{-width_1 \leq t\leq width_2} f(x+t), 前回の幅、今回の幅
常に、区間上で最小値を持ち傾きが1ずつ変わる凸な関数であることが維持される。（区間は1点かも）
傾きが変わる点の集合S_f = S_f_lower + S_f_upperを持っていく。
S_f_lower, S_upperは一斉に定数を足す：変化量のみ持つ
"""

N = int(input())
LR = [[int(x) for x in input().split()] for _ in range(N)]

# initialize
L,R = LR[0]
S_lower = [-L]
S_upper = [L]
min_f = 0
add_lower = 0
add_upper = 0
prev_w = R - L

push_L = lambda x: heappush(S_lower, -x)
push_R = lambda x: heappush(S_upper, x)
pop_L = lambda: -heappop(S_lower)
pop_R = lambda: heappop(S_upper)

for L,R in LR[1:]:
    w = R - L
    # 平行移動とのminをとるステップ
    add_lower -= w
    add_upper += prev_w
    # abs(x-L) を加えるステップ
    # abs は瞬間に2傾きが変わるので
    x = pop_L() + add_lower
    y = pop_R() + add_upper
    a,b,c,d = sorted([x,y,L,L])
    push_L(a - add_lower)
    push_L(b - add_lower)
    push_R(c - add_upper)
    push_R(d - add_upper)
    min_f += c-b
    prev_w = w

print(min_f)