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πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
France
Joined March 2021
139
Following
40
Followers
46
Posts
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
7 months ago
@n0dyas
Medft25
Β retweeted
Shirin
@Shirink_13
about 1 year ago
Hind Rajab was 6 and she was shot with 335 bullets.
Medft25
Β retweeted
Hassan Mafi β
@thatdayin1992
about 1 year ago
btw, according to Charlie Kirk's logic, his death is his own fault, not the shooter's.
Medft25
Β retweeted
G
@pasderalentir
about 1 year ago
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@lorisbgr
@slwhrr
Nice try diddy
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@slwhrr
@lorisbgr
Tu me parles pas toi ...
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@lorisbgr
C tellement prΓ©visible qu'ils seront pas postΓ©s Γ 16h
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@slwhrr
Donner une crise cardiaque Γ qqn c'est lΓ©gal ouai
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@slwhrr
Vzy Γ§a commence les jumpscares chaque 5 minutes
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@Vlnd06
Mdr je me demandais Γ§a moi aussi quand je l'ai vu
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@sbaWAEL
Medft25's tweet gif.
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@matthlan
En vrai met les on sait jamais si tu tombes sur un examinateur chiant il peut te questionner dessus
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@lucazaco Pardon pour le jumpscare.. c'est demain
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@martuun_
Je suis d'origines schrΓΆdingeriennes
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
Les serveurs ccinp Γ partir de ce soir
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@FlixM03
0 vanne le premier jallon commence en janvier de l'annΓ©e de sup
#0honte
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@antoinzzzz
En vrai fait le asap pcq c sΓ»r que le site va buger mardi (ptete mΓͺme demain)
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
@marmarjk97
Askip ils ont vu Γ§a
πβ(βπ/βπ‘) =β(βΒ²/2π)Ξπ+π(π₯, π‘)π
@Medft25
over 1 year ago
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